{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Cho $\\triangle{MNP}$, c\u00e1c \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $MD; NE$ c\u1eaft nhau \u1edf $I$. Trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau kh\u1eb3ng \u0111\u1ecbnh n\u00e0o \u0111\u00fang? ","select":[" <span class='basic_left'> A. \u0110i\u1ec3m $I$ c\u00e1ch \u0111\u1ec1u $3$ \u0111\u1ec9nh c\u1ee7a tam gi\u00e1c <\/span> ","<span class='basic_left'> B. \u0110i\u1ec3m $I$ c\u00e1ch \u0111\u1ec1u $3$ c\u1ea1nh c\u1ee7a tam gi\u00e1c <\/span>","<span class='basic_left'> C. $MI = \\dfrac{2}{3}MD$ <\/span> "],"hint":"","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D01.png' \/><\/center> <br\/> Theo t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c th\u00ec: <br\/> Ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u00f9ng \u0111i qua m\u1ed9t \u0111i\u1ec3m. \u0110i\u1ec3m n\u00e0y c\u00e1ch \u0111\u1ec1u ba c\u1ea1nh c\u1ee7a tam gi\u00e1c \u0111\u00f3 <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B <\/span>","column":1}]}],"id_ques":1921},{"time":24,"part":[{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["2","3","4"]],"list":[{"point":5,"img":"","ques":"Ch\u1ecdn c\u00e1c kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau:","hint":"","column":1,"number_true":3,"select":["A. \u0110i\u1ec3m c\u00e1ch \u0111\u1ec1u hai \u0111\u01b0\u1eddng th\u1eb3ng ch\u1ee9a hai c\u1ea1nh c\u1ee7a m\u1ed9t g\u00f3c th\u00ec n\u1eb1m tr\u00ean tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c \u0111\u00f3.","B. \u0110i\u1ec3m n\u1eb1m trong tam gi\u00e1c v\u00e0 c\u00e1ch \u0111\u1ec1u ba c\u1ea1nh c\u1ee7a tam gi\u00e1c l\u00e0 giao \u0111i\u1ec3m ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c \u0111\u00f3","C. \u0110i\u1ec3m c\u00e1ch \u0111\u1ec1u ba \u0111\u01b0\u1eddng th\u1eb3ng ch\u1ee9a ba c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c lu\u00f4n n\u1eb1m trong tam gi\u00e1c \u0111\u00f3.","D. \u0110i\u1ec3m n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c th\u00ec c\u00e1ch \u0111\u1ec1u hai c\u1ea1nh c\u1ee7a g\u00f3c "],"explain":"<span class='basic_left'> A - SAI theo t\u00ednh ch\u1ea5t tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c (\u0111i\u1ec3m \u0111\u00f3 ph\u1ea3i n\u1eb1m b\u00ean trong g\u00f3c) <br\/> B - \u0110\u00daNG theo t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong tam gi\u00e1c <br\/> C - \u0110\u00daNG <br\/> D - \u0110\u00daNG theo t\u00ednh ch\u1ea5t tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c <\/span>"}]}],"id_ques":1922},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong tam gi\u00e1c $DEF$ \u0111\u1ec1u. Trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sai? ","select":["<span class='basic_left'> A. Giao \u0111i\u1ec3m $DI$ v\u1edbi $EF$ l\u00e0 $M$ th\u00ec $DM$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c v\u00e0 l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn \u1ee9ng v\u1edbi c\u1ea1nh $EF$ <\/span> ","<span class='basic_left'> B. G\u1ecdi giao \u0111i\u1ec3m $DI$ v\u1edbi $EF$ l\u00e0 $M$ th\u00ec $IM = \\dfrac{2}{3}DM$ <\/span>","<span class='basic_left'> C. $I$ l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c $DEF$ <\/span>","<span class='basic_left'> D. $I$ c\u00e1ch \u0111\u1ec1u $3$ c\u1ea1nh c\u1ee7a tam gi\u00e1c <\/span>"],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> A - Ch\u1ee9ng minh $EM = FM$ b\u1eb1ng c\u00e1ch x\u00e9t $\\triangle{FDM}$ v\u00e0 $\\triangle{EDM}$ <br\/> B - Ch\u1ec9 ra \u0111\u00e2u l\u00e0 tr\u1ecdng t\u00e2m v\u00e0 d\u1ef1a v\u00e0o t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D02.png' \/><\/center> <br\/> $\\blacktriangleright$ A - \u0110\u00fang v\u00ec: <br\/> X\u00e9t $\\triangle{FDM}$ v\u00e0 $\\triangle{EDM}$ c\u00f3: <br\/> $\\begin{cases} DF = DE (\\triangle{DEF} \\hspace{0,2cm} \\text{\u0111\u1ec1u}) \\\\ \\widehat{FDM} = \\widehat{EDM} (gt) \\\\ DM \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{FDM} = \\widehat{EDM}$ (c.g.c) <br\/> $\\Rightarrow$ $EM = FM$ hay $DM$ l\u00e0 trung tuy\u1ebfn \u1ee9ng v\u1edbi c\u1ea1nh $EF$ <br\/> $\\blacktriangleright$ SAI v\u00ec: <br\/> Theo c\u00e2u tr\u00ean $DM$ l\u00e0 trung tuy\u1ebfn <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u00f3 $FI$ v\u00e0 $EI$ c\u0169ng l\u00e0 trung tuy\u1ebfn c\u1ee7a tam gi\u00e1c $DEF$ <br\/> $\\Rightarrow$ $I$ l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c $DEF$ <br\/> Theo t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m: $DI = \\dfrac{2}{3}DM$ $\\Rightarrow$ $IM = \\dfrac{1}{3}DM$ <br\/> $\\blacktriangleright$ C - \u0110\u00fang (cmt) <br\/> $\\blacktriangleright$ D - \u0110\u00fang theo t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong tam gi\u00e1c <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n sai l\u00e0: B <\/span>","column":1}]}],"id_ques":1923},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Cho h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D03.png' \/><\/center> <br\/> Bi\u1ebft $\\widehat{KOH} = 28^{o}$; $\\widehat{IOH} = 28^{o}$; $IH = 12cm$. $KH$ b\u1eb1ng bao nhi\u00eau x\u0103ng ti m\u00e9t? ","select":["A. $KH = 10cm$ ","B. $KH = 8cm$ ","C. $KH = 12cm$ ","D. $KH = 15cm$ "],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $OH$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $KOI$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> D\u1ef1a v\u00e0o t\u00ednh ch\u1ea5t tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c \u0111\u1ec3 t\u00ednh $KH$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D03.png' \/><\/center> <br\/> $\\blacktriangleright$ Theo \u0111\u1ec1 b\u00e0i ta c\u00f3: $\\widehat{KOH} = \\widehat{IOH} = 28^{o}$ <br\/> $\\Rightarrow$ $OH$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{KOI}$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $\\blacktriangleright$ M\u1eb7t kh\u00e1c, v\u00ec $H$ thu\u1ed9c tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{KOI}$ <br\/> N\u00ean theo t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c th\u00ec $HK = HI = 12cm$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: Ngo\u00e0i c\u00e1ch l\u00e0m tr\u00ean ta c\u00f2n c\u00f3 th\u1ec3 ch\u1ee9ng minh hai tam gi\u00e1c b\u1eb1ng nhau <br\/> X\u00e9t hai tam gi\u00e1c vu\u00f4ng $OKH$ v\u00e0 $OIH$ c\u00f3: <br\/> $OH$ chung <br\/> $\\widehat{KOH} = \\widehat{IOH}$ (gt) <br\/> $\\Rightarrow$ $\\triangle{OKH} = \\triangle{OIHI}$ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow$ $KH = IH = 12 (cm)$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <\/i> <\/span> ","column":2}]}],"id_ques":1924},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Cho h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D04.png' \/><\/center> <br\/> Bi\u1ebft $BO; CO$ l\u00e0 hai tia ph\u00e2n gi\u00e1c \u1ee9ng v\u1edbi $\\widehat{B}$ v\u00e0 $\\widehat{C}$. T\u00ednh $\\widehat{BOC}$. ","select":["A. $\\widehat{BOC} = 50^{o}$ ","B. $\\widehat{BOC} = 100^{o}$ ","C. $\\widehat{BOC} = 130^{o}$ ","D. $\\widehat{BOC} = 80^{o}$ "],"hint":"D\u1ef1a v\u00e0o \u0111\u1ecbnh ngh\u0129a tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c v\u00e0 \u0111\u1ecbnh l\u00fd t\u1ed5ng ba g\u00f3c trong m\u1ed9t tam gi\u00e1c","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $\\widehat{B} + \\widehat{C}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\widehat{OBC} + \\widehat{OCB}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh $\\widehat{BOC}$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D04A.png' \/><\/center> <br\/> $\\blacktriangleright$ \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd t\u1ed5ng ba g\u00f3c trong tam gi\u00e1c cho $\\triangle{ABC}$, ta c\u00f3: <br\/> $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o}$ <br\/> $\\begin{align}\\Rightarrow \\widehat{B} + \\widehat{C} &= 180^{o} - \\widehat{A} \\\\ &= 180^{o} - 80^{o} \\\\ &= 100^{o} \\end{align}$ <br\/> $\\blacktriangleright$ Trong $\\triangle{BOC}$ c\u00f3: <br\/> $\\widehat{B_{1}} + \\widehat{C_{1}} + \\widehat{O_{1}} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong tam gi\u00e1c $BOC$) <br\/> M\u00e0 $\\widehat{B_{1}} + \\widehat{C_{1}} = \\dfrac{\\widehat{B} + \\widehat{C}}{2} = \\dfrac{100^{o}}{2} = 50^{o}$ (V\u00ec $BO; CO$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{B}$ v\u00e0 $\\widehat{C}$) <br\/> $\\Rightarrow$ $\\widehat{O_{1}} = 180^{o} - (\\widehat{B_{1}} + \\widehat{C_{1}}) = 180^{o} - 50^{o} = 130^{o}$ <br\/> $\\Rightarrow$ $\\widehat{BOC} = 130^{o}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C <\/span>","column":2}]}],"id_ques":1925},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho tam gi\u00e1c $MNP$. Tr\u00ean tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $N$ l\u1ea5y \u0111i\u1ec3m $I$ n\u1eb1m trong tam gi\u00e1c $MNP$ sao cho $I$ c\u00e1ch \u0111\u1ec1u hai c\u1ea1nh $MN$ v\u00e0 $MP$. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y sai? ","select":[" <span class='basic_left'> A. \u0110i\u1ec3m $I$ kh\u00f4ng n\u1eb1m tr\u00ean tia ph\u00e2n gi\u00e1c g\u00f3c $P$ <\/span> ","<span class='basic_left'> B. \u0110i\u1ec3m $I$ c\u00e1ch \u0111\u1ec1u $MN$ v\u00e0 $NP$ <\/span>","<span class='basic_left'> C. \u0110i\u1ec3m $I$ c\u00e1ch \u0111\u1ec1u $MN; MP; NP$ <\/span>","<span class='basic_left'> D. \u0110i\u1ec3m $I$ n\u1eb1m tr\u00ean tia ph\u00e2n gi\u00e1c g\u00f3c $M$ <\/span> "],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ec9 ra $I$ thu\u1ed9c tia ph\u00e2n gi\u00e1c g\u00f3c $M$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ec9 ra $I$ l\u00e0 giao \u0111i\u1ec3m $3$ tia ph\u00e2n gi\u00e1c trong tam gi\u00e1c <br\/> <b> B\u01b0\u1edbc 3: <\/b>Ch\u1ecdn kh\u1eb3ng \u0111\u1ecbnh sai. <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D05.png' \/><\/center> <br\/> $\\blacktriangleright$ Theo gi\u1ea3 thi\u1ebft, $I$ thu\u1ed9c tia ph\u00e2n gi\u00e1c g\u00f3c $N$ (1) <br\/> M\u1eb7t kh\u00e1c $IH = IK$ (gt) $\\Rightarrow$ $I$ thu\u1ed9c tia ph\u00e2n gi\u00e1c g\u00f3c $M$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $I$ thu\u1ed9c tia ph\u00e2n gi\u00e1c g\u00f3c $P$ (t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c) <br\/> $\\Rightarrow$ $I$ c\u00e1ch \u0111\u1ec1u ba c\u1ea1nh c\u1ee7a tam gi\u00e1c <br\/> <span class='basic_pink'> V\u1eady kh\u1eb3ng \u0111\u1ecbnh sai l\u00e0: A <\/span>","column":1}]}],"id_ques":1926},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["135"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"Cho tam gi\u00e1c $DEF$ c\u00f3 $\\widehat{D} = \\widehat{E} + \\widehat{F}$. Hai \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $\\widehat{E}$ v\u00e0 $\\widehat{F}$ c\u1eaft nhau t\u1ea1i $O$. T\u00ednh $\\widehat{EOF}$. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $\\widehat{EOF} =$ _input_$^{o}$ ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $\\widehat{E} + \\widehat{F}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\dfrac{\\widehat{E} + \\widehat{F}}{2}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh $\\widehat{EOF}$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D06.png' \/><\/center> $\\blacktriangleright$ $\\triangle{DEF}$ c\u00f3: <br\/> $\\widehat{D} + \\widehat{E} + \\widehat{F} = 180^{o}$ (t\u1ed5ng $3$ g\u00f3c trong $1$ tam gi\u00e1c) (1) <br\/> M\u1eb7t kh\u00e1c ta c\u00f3: $\\widehat{D} = \\widehat{E} + \\widehat{F}$ (gt) <br\/> $\\Rightarrow$ $\\widehat{D} + \\widehat{D} = 180^{o}$ $\\Rightarrow$ $\\widehat{D} = 90^{o}$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $\\widehat{E} + \\widehat{F} = 180^{o} - 90^{o} = 90^{o}$ <br\/> $\\blacktriangleright$ Trong $\\triangle{EOF}$ c\u00f3: <br\/> $\\widehat{E_{1}} + \\widehat{O_{1}} + \\widehat{F_{1}} = 180^{o}$ (t\u1ed5ng $3$ g\u00f3c trong $1$ tam gi\u00e1c) (3) <br\/> M\u00e0 $\\widehat{E_{1}} + \\widehat{F_{1}} = \\dfrac{\\widehat{E} + \\widehat{F}}{2} = \\dfrac{90^{o}}{2} = 45^{o}$ ($EO$; $FO$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{E}$ v\u00e0 $\\widehat{F}$) (4) <br\/> T\u1eeb (3) v\u00e0 (4) $\\Rightarrow$ $\\widehat{O_{1}} = 180^{o} - 45^{o} = 135^{o}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p c\u1ea7n \u0111i\u1ec1n l\u00e0: $135$ <\/span>"}]}],"id_ques":1927},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["22"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"Cho h\u00ecnh v\u1ebd, t\u00ednh $\\widehat{CBO}$. <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D07.png' \/><\/center> <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $\\widehat{CBO} =$ _input_$^{o}$ ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ec9 ra $BO$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{B}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> D\u1ef1a v\u00e0o \u0111\u1ecbnh ngh\u0129a tia ph\u00e2n gi\u00e1c \u0111\u1ec3 t\u00ednh $\\widehat{CBO}$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D07.png' \/><\/center> Ta th\u1ea5y $O$ c\u00e1ch \u0111\u1ec1u ba c\u1ea1nh c\u1ee7a tam gi\u00e1c <br\/> $\\Rightarrow$ $O$ l\u00e0 giao \u0111i\u1ec3m ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c $ABC$ <br\/> $\\Rightarrow$ $BO$ l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c $B$ <br\/> $\\Rightarrow$ $\\widehat{CBO} = \\widehat{ABO} = \\dfrac{\\widehat{B}}{2} = \\dfrac{44^{o}}{2} = 22^{o}$ (\u0111\u1ecbnh ngh\u0129a tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c) <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p c\u1ea7n \u0111i\u1ec1n l\u00e0: $22$ <\/span>"}]}],"id_ques":1928},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["11"],["11"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"Cho h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D08.png' \/><\/center> <br\/> Bi\u1ebft $IP = 11cm$. T\u00ednh $IM; IN$. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $IM = $ _input_ $cm$; $IN = $ _input_ $cm$","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ec9 ra $I$ l\u00e0 giao \u0111i\u1ec3m c\u00e1c \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c $DEF$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $IM; IN$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D08.png' \/><\/center> <br\/> Ta c\u00f3: <br\/> $\\widehat{DEI} = \\widehat{IEF}$ (gt) $\\Rightarrow$ $EI$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{E}$ (\u0111\u1ecbnh ngh\u0129a) (1) <br\/> $\\widehat{DFI} = \\widehat{EFI}$ (gt) $\\Rightarrow$ $FI$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{F}$ (\u0111\u1ecbnh ngh\u0129a) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $I$ thu\u1ed9c tia ph\u00e2n gi\u00e1c $\\widehat{D}$ (ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1eaft nhau t\u1ea1i m\u1ed9t \u0111i\u1ec3m) <br\/> $\\Rightarrow$ $I$ l\u00e0 giao \u0111i\u1ec3m $3$ \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c $DEF$ <br\/> $\\Rightarrow$ $I$ c\u00e1ch \u0111\u1ec1u $3$ c\u1ea1nh c\u1ee7a tam gi\u00e1c hay $IM = IN = IP = 11cm$ (t\u00ednh ch\u1ea5t giao \u0111i\u1ec3m ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c) <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p c\u1ea7n \u0111i\u1ec1n l\u00e0: $11$ v\u00e0 $11$ <\/span>"}]}],"id_ques":1929},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["28"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"Cho h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D09.png' \/><\/center> <br\/> T\u00ednh $\\widehat{KHO}$. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $\\widehat{KHO} =$ _input_$^{o}$ ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $HO$ l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c $\\widehat{H}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\widehat{KHO}$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D09.png' \/><\/center> <br\/> $\\blacktriangleright$ Ta c\u00f3: $\\widehat{HIO} = \\widehat{KIO}$ (gt) $\\Rightarrow$ $IO$ l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c $I$ (1) <br\/> $\\widehat{IKO} = \\widehat{HKO}$ (gt) $\\Rightarrow$ $KO$ l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c $K$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $O$ l\u00e0 giao \u0111i\u1ec3m hai tia ph\u00e2n gi\u00e1c <br\/> $\\Rightarrow$ $O$ thu\u1ed9c tia ph\u00e2n gi\u00e1c g\u00f3c $H$ (t\u00ednh ch\u1ea5t $3$ \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong tam gi\u00e1c) <br\/> $\\Rightarrow$ $\\widehat{IHO} = \\widehat{KHO} = \\dfrac{\\widehat{H}}{2} = \\dfrac{56^{o}}{2} = 28^{o}$ (\u0111\u1ecbnh ngh\u0129a \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c) <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p c\u1ea7n \u0111i\u1ec1n l\u00e0: $28$ <\/span>"}]}],"id_ques":1930},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["40"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"Cho h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D10.png' \/><\/center> <br\/> T\u00ednh $\\widehat{LMO}$. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $\\widehat{LMO} =$ _input_$^{o}$ ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh g\u00f3c $M$ sau \u0111\u00f3 ch\u1ee9ng minh $MO$ l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c $M$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\widehat{LMO}$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D10.png' \/><\/center> <br\/> $\\blacktriangleright$ Ta c\u00f3: $\\widehat{I_{1}} + \\widehat{L_{1}} + \\widehat{IOL} = 180^{o}$ <br\/> $ \\begin{align} \\Rightarrow \\widehat{I_{1}} + \\widehat{L_{1}} &= 180^{o} - \\widehat{IOL} \\\\ &= 180^{o} - 130^{o} \\\\ &= 50^{o} (1) \\end{align}$ <br\/> M\u1eb7t kh\u00e1c ta c\u00f3: $\\widehat{MIO} = \\widehat{OIL}$ (gt) (2) <br\/> $\\widehat{ILO} = \\widehat{MLO}$ (gt) (3) <br\/> T\u1eeb (1), (2), (3) $\\Rightarrow$ $\\widehat{I} + \\widehat{L} = 2. (\\widehat{OIL} + \\widehat{ILO}) = 2 . 50^{o} = 100^{o}$ <br\/> $\\blacktriangleright$ Trong $\\triangle{IML}$ c\u00f3: <br\/> $\\widehat{M} + \\widehat{L} + \\widehat{I} = 180^{o}$ <br\/> $\\begin{align} \\Rightarrow \\widehat{M} &= 180^{o} - (\\widehat{I} + \\widehat{L}) \\\\ &= 180^{o} - 100^{o} \\\\ &= 80^{o} \\end{align}$ <br\/> $\\blacktriangleright$ M\u1eb7t kh\u00e1c t\u1eeb (2) $\\Rightarrow$ $IO$ l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c $I$ <br\/> T\u1eeb (3) $\\Rightarrow$ $LO$ l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c $L$ <br\/> $\\Rightarrow$ $O$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a tia ph\u00e2n gi\u00e1c g\u00f3c $I$ v\u00e0 g\u00f3c $L$ <br\/> $\\Rightarrow$ $O$ thu\u1ed9c tia ph\u00e2n gi\u00e1c g\u00f3c $M$ <br\/> $\\Rightarrow$ $\\widehat{LMO} = \\dfrac{\\widehat{M}}{2} = \\dfrac{80^{o}}{2} = 40^{o}$ (\u0111\u1ecbnh ngh\u0129a \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c) <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p c\u1ea7n \u0111i\u1ec1n l\u00e0: $40$ <\/span>"}]}],"id_ques":1931},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Cho h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D11.png' \/><\/center> <br\/> T\u00ednh $\\widehat{D_{1}}$. ","select":["A. $\\widehat{D_{1}} = 40^{o} $ ","B. $\\widehat{D_{1}} = 20^{o} $","C. $\\widehat{D_{1}} = 95^{o} $ ","D. $\\widehat{D_{1}} = 85^{o} $ "],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $\\widehat{C}$ sau \u0111\u00f3 t\u00ednh $\\widehat{ACD}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\widehat{D_{1}}$ d\u1ef1a v\u00e0o t\u1ed5ng ba g\u00f3c trong tam gi\u00e1c $ACD$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D11.png' \/><\/center> <br\/> $\\blacktriangleright$ \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd t\u1ed5ng ba g\u00f3c trong $\\triangle{ABC}$, ta c\u00f3: <br\/> $\\begin{align} \\widehat{A} + \\widehat{B} + \\widehat{C} &= 180^{o} \\\\ \\widehat{C} &= 180^{o} - (\\widehat{B} + \\widehat{A}) \\\\ &= 180^{o} - (75^{o} + 65^{o}) \\\\ &= 180^{o} - 140^{o} \\\\ &= 40^{o} \\end{align}$ <br\/> V\u00ec $\\widehat{ACD} = \\widehat{BCD} = \\dfrac{\\widehat{C}}{2} = \\dfrac{40^{o}}{2} = 20^{o}$ (\u0111\u1ecbnh ngh\u0129a \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c) <br\/> $\\blacktriangleright$ Trong $\\triangle{ACD}$ c\u00f3: <br\/> $\\widehat{A} + \\widehat{ACD} + \\widehat{D_{1}} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong m\u1ed9t tam gi\u00e1c) <br\/> $\\begin{align} \\Rightarrow \\widehat{D_{1}} &= 180^{o} - (\\widehat{A} + \\widehat{ACD}) \\\\ &= 180^{o} - (75^{o} + 20^{o}) \\\\ &= 85^{o} \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span>","column":2}]}],"id_ques":1932},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D12.png' \/><\/center> <br\/> T\u00ednh $\\widehat{PSQ}$. ","select":["A. $\\widehat{PSQ} = 50^{o} $ ","B. $\\widehat{PSQ} = 60^{o} $","C. $\\widehat{PSQ} = 30^{o} $ ","D. $\\widehat{PSQ} = 55^{o} $ "],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $\\widehat{P}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\widehat{PSQ}$ d\u1ef1a v\u00e0o g\u00f3c ngo\u00e0i $\\widehat{PQx}$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D12.png' \/><\/center> <br\/> $\\blacktriangleright$ V\u00ec $\\widehat{QPR} = \\widehat{SPR}$ (gt) <br\/> $\\Rightarrow$ $\\widehat{P} = 2.\\widehat{SPR} = 2 . 30^{o} = 60^{o}$ <br\/> $\\blacktriangleright$ M\u1eb7t kh\u00e1c ta c\u00f3: $\\widehat{PQx} = \\widehat{P} + \\widehat{S}$ (\u0111\u1ecbnh l\u00fd g\u00f3c ngo\u00e0i) <br\/> $ \\begin{align} \\Rightarrow \\widehat{S} &= \\widehat{PQx} - \\widehat{P} \\\\ &= 110^{o} - 60^{o} \\\\ &= 50^{o} \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A <\/span>","column":2}]}],"id_ques":1933},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["50"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{B} = 60^{o}; \\widehat{C} = 40^{o}$. T\u00ednh g\u00f3c gi\u1eefa \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BAC}$ v\u00e0 \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c k\u1ebb t\u1eeb $B$ \u0111\u1ebfn $AC$. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> _input_$^{o}$ ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $\\widehat{BAC}$ r\u1ed3i t\u00ednh $\\widehat{CAM}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\widehat{AIH}$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D13.png' \/><\/center> <br\/> G\u1ecdi $AM$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c g\u00f3c $A$ <br\/> $BH \\perp AC$; $BH$ c\u1eaft $AM$ t\u1ea1i $I$ <br\/> $\\blacktriangleright$ Trong $\\triangle{ABC}$ c\u00f3: <br\/> $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong $1$ tam gi\u00e1c) <br\/> $ \\begin{align} \\Rightarrow \\widehat{A} &= 180^{o} - ( \\widehat{B} + \\widehat{C}) \\\\ &= 180^{o} - (60^{o} + 40^{o}) \\\\ &= 180^{o} - 100^{o} \\\\ &= 80^{o} \\end{align}$ <br\/> V\u00ec $AM$ l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c $A$ n\u00ean: $\\widehat{BAM} = \\widehat{CAM} = \\dfrac{\\widehat{A}}{2} = \\dfrac{80^{o}}{2} = 40^{o}$ (\u0111\u1ecbnh ngh\u0129a \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c) <br\/> $\\blacktriangleright$ Trong $\\triangle{AIH}$ c\u00f3: <br\/> $\\widehat{I_{1}} + \\widehat{IAH} + \\widehat{AHI} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong $1$ tam gi\u00e1c) <br\/> $\\begin{align} \\Rightarrow \\widehat{I_{1}} &= 180^{o} - \\widehat{IAH} - \\widehat{AHI} \\\\ &= 180^{o} - 40^{o} - 90^{o} \\\\ &= 50^{o} \\end{align}$ <br\/> V\u1eady g\u00f3c gi\u1eefa \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BAC}$ v\u00e0 \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c k\u1ebb t\u1eeb $B$ \u0111\u1ebfn $AC$ l\u00e0 $50^{o}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p c\u1ea7n \u0111i\u1ec1n l\u00e0: $50$ <\/span>"}]}],"id_ques":1934},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho tam gi\u00e1c $ABC$ n\u1ebfu \u0111\u01b0\u1eddng trung tuy\u1ebfn $AM$ \u0111\u1ed3ng th\u1eddi l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c g\u00f3c $A$ th\u00ec tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG ","B. SAI "],"hint":"K\u1ebb $MH \\perp AB; MK \\perp AC$","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $\\widehat{B} = \\widehat{C}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ec9 ra tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D14.png' \/><\/center> <br\/> $\\blacktriangleright$ K\u1ebb $MH \\perp AB; MK \\perp AC$ <br\/> V\u00ec $AM$ l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c $A$ n\u00ean $MK = MH$ (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c) <br\/> X\u00e9t hai tam gi\u00e1c vu\u00f4ng $HBM$ v\u00e0 $KCM$ c\u00f3: <br\/> $\\begin{cases} BM = CM (gt) \\\\ MH = MK (cmt) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{HBM} = \\triangle{KCM}$ (c\u1ea1nh huy\u1ec1n - c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow$ $\\widehat{B} = \\widehat{C}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow$ $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span>","column":2}]}],"id_ques":1935},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Cho tam gi\u00e1c $DEF$, c\u00e1c \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $EM; FN$ c\u1eaft nhau t\u1ea1i $O$. Khi \u0111\u00f3 $\\widehat{EOF}$ l\u00e0 g\u00f3c nh\u1ecdn. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG ","B. SAI "],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> D\u1ef1a v\u00e0o \u0111\u1ecbnh l\u00fd t\u1ed5ng ba g\u00f3c t\u00ednh $\\widehat{E} + \\widehat{F}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> D\u1ef1a v\u00e0o \u0111\u1ecbnh ngh\u0129a \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c t\u00ednh $\\widehat{E_{1}} + \\widehat{F_{1}}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh $\\widehat{EOF}$ d\u1ef1a v\u00e0o \u0111\u1ecbnh l\u00fd t\u1ed5ng ba g\u00f3c trong tam gi\u00e1c $EOF$, bi\u1ebfn \u0111\u1ed5i \u0111\u1ec3 so s\u00e1nh $\\widehat{EOF}$ v\u1edbi $90^{o}$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D15.png' \/><\/center> <br\/> $\\blacktriangleright$ $\\triangle{DEF}$ c\u00f3: $\\widehat{D} + \\widehat{E} + \\widehat{F} = 180^{o}$ (t\u1ed5ng $3$ g\u00f3c trong $1$ tam gi\u00e1c) <br\/> $\\Rightarrow$ $\\widehat{D} = 180^{o} - (\\widehat{E} + \\widehat{F})$ <br\/> V\u00ec $EM; FN$ l\u00e0 c\u00e1c \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c n\u00ean ta c\u00f3: <br\/> $\\widehat{E_{1}} = \\dfrac{\\widehat{E}}{2}$; $\\widehat{F_{1}} = \\dfrac{\\widehat{F}}{2}$ (\u0111\u1ecbnh ngh\u0129a \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c) <br\/> $\\blacktriangleright$ Trong $\\triangle{EOF}$ c\u00f3: $\\widehat{O_{1}} + \\widehat{E_{1}} + \\widehat{F_{1}} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong $1$ tam gi\u00e1c) <br\/> $ \\begin{align} \\Rightarrow \\widehat{O_{1}} &= 180^{o} - (\\widehat{E_{1}} + \\widehat{F_{1}}) \\\\ &= 180^{o} - \\left( \\dfrac{\\widehat{E}}{2} + \\dfrac{\\widehat{F}}{2} \\right) \\\\ &= 180^{o} - \\dfrac{1}{2}(\\widehat{E} + \\widehat{F}) \\\\ &= 180^{o} - \\dfrac{1}{2}. (180^{o} - \\widehat{D}) \\\\ &= 90^{o} + \\dfrac{\\widehat{D}}{2} > 90^{o} \\end{align} $ <br\/> Suy ra, $\\widehat{EOF}$ l\u00e0 g\u00f3c t\u00f9 <br\/> <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> SAI <\/span>","column":2}]}],"id_ques":1936},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Cho h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D16.png' \/><\/center> <br\/> Bi\u1ebft $\\widehat{BAC} = 50^{o}$; $\\widehat{ABC} = 60^{o}$. Khi \u0111\u00f3 s\u1ed1 \u0111o c\u1ee7a g\u00f3c $BIC$ l\u00e0: ","select":["A. $20^{o} $ ","B. $ 25^{o} $","C. $ 30^{o} $ ","D. $ 35^{o} $ "],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $\\widehat{ACB}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\widehat{CBI}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh $\\widehat{ACI}$ r\u1ed3i t\u00ednh $\\widehat{BCI}$ <br\/> <b> B\u01b0\u1edbc 4: <\/b> T\u00ednh $\\widehat{BIC}$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D16.png' \/><\/center> <br\/> $\\blacktriangleright$ $\\triangle{ABC}$ c\u00f3: $\\widehat{ABC} + \\widehat{ACB} + \\widehat{BAC} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong $1$ tam gi\u00e1c) <br\/> $ \\begin{align} \\Rightarrow \\widehat{ACB} &= 180^{o} - (\\widehat{ABC} + \\widehat{BAC}) \\\\ &= 180^{o} - (50^{o} + 60^{o}) \\\\ &= 180^{o} - 110^{o} \\\\ &= 70^{o} \\end{align}$ <br\/> $\\blacktriangleright$ V\u00ec $\\widehat{ACB}$ v\u00e0 $\\widehat{ACD}$ l\u00e0 hai g\u00f3c k\u1ec1 b\u00f9 n\u00ean: $\\widehat{ACD} = 180^{o} - \\widehat{ACB} = 180^{o} - 70^{o} = 110^{o}$ <br\/> $\\blacktriangleright$ V\u00ec $\\widehat{ABI} = \\widehat{CBI}$ (gt) $\\Rightarrow$ $\\widehat{ABI} = \\widehat{CBI} = \\dfrac{\\widehat{ABC}}{2} = \\dfrac{60^{o}}{2} = 30^{o}$ <br\/> $\\blacktriangleright$ Ta c\u00f3: $\\widehat{ACI} = \\widehat{DCI}$ (gt) $\\Rightarrow$ $\\widehat{ACI} = \\widehat{DCI} = \\dfrac{\\widehat{ACD}}{2} = \\dfrac{110^{o}}{2} = 55^{o}$ <br\/> M\u1eb7t kh\u00e1c ta c\u00f3: $\\widehat{BCI} = \\widehat{BCA} + \\widehat{ACI} = 70^{o} + 55^{o} = 125^{o}$ <br\/> $\\blacktriangleright$ Trong $\\triangle{BIC}$ c\u00f3: $\\widehat{CBI} + \\widehat{BCI} + \\widehat{BIC} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong $1$ tam gi\u00e1c) <br\/> $\\begin{align} \\Rightarrow \\widehat{BIC} &= 180^{o} - (\\widehat{CBI} + \\widehat{BCI}) \\\\ &= 180^{o} - (30^{o} + 125^{o}) \\\\ &= 180^{o} - 155^{o} \\\\ &= 25^{o} \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B <\/span>","column":2}]}],"id_ques":1937},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$. Tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $BC$ kh\u00f4ng ch\u1ee9a $A$, d\u1ef1ng $\\triangle{BCD}$ vu\u00f4ng c\u00e2n t\u1ea1i $D$. Khi \u0111\u00f3 $AD$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $A$. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG ","B. SAI "],"hint":"V\u1eadn d\u1ee5ng \u0111\u1ecbnh l\u00fd: \u0110i\u1ec3m n\u1eb1m b\u00ean trong m\u1ed9t g\u00f3c v\u00e0 c\u00e1ch \u0111\u1ec1u hai c\u1ea1nh c\u1ee7a g\u00f3c th\u00ec n\u1eb1m tr\u00ean tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c \u0111\u00f3. ","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> K\u1ebb $DI \\perp AB; DH \\perp AC$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $DI = DH$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D17.png' \/><\/center> <br\/> $\\blacktriangleright$ K\u1ebb $DI \\perp AB$, v\u00ec $AB \\perp AC$ (gt) $\\Rightarrow$ $DI \/\/ AC$ <br\/> K\u1ebb $DH \\perp AC$ $\\Rightarrow$ $DH \\perp DI$ <br\/> $\\blacktriangleright$ X\u00e9t hai tam gi\u00e1c vu\u00f4ng $BDI$ v\u00e0 $CDH$ c\u00f3: <br\/> $\\begin{cases} BD = DC (gt) \\\\ \\widehat{D_{1}} = \\widehat{D_{3}} (\\text{c\u00f9ng} \\hspace{0,2cm} \\text{ph\u1ee5} \\hspace{0,2cm} \\text{v\u1edbi} \\hspace{0,2cm} \\widehat{BDH}) \\end{cases}$ <br\/> $\\Rightarrow$ $\\widehat{BDI} = \\widehat{CDH}$ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow$ $DI = DH$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow$ $AD$ l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c $A$ (t\u00ednh ch\u1ea5t tia ph\u00e2n gi\u00e1c) <br\/> <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span>","column":2}]}],"id_ques":1938},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$. Tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $AB$ kh\u00f4ng ch\u1ee9a \u0111i\u1ec3m $C$ v\u1ebd tia $Ax$, tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $AC$ kh\u00f4ng ch\u1ee9a \u0111i\u1ec3m $B$ v\u1ebd tia $Ay$ so cho $\\widehat{xAB} = \\widehat{yAC}$. K\u1ebb $BD \\perp Ax$ $(D \\in Ax)$, $CE \\perp Ay$ $(E \\in Ay)$. \u0110\u01b0\u1eddng th\u1eb3ng $BD$ v\u00e0 $CE$ c\u1eaft nhau t\u1ea1i $K$. Khi \u0111\u00f3 $KA$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{DKE}$. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG ","B. SAI "],"hint":"V\u1eadn d\u1ee5ng \u0111\u1ecbnh l\u00fd: \u0110i\u1ec3m n\u1eb1m b\u00ean trong m\u1ed9t g\u00f3c v\u00e0 c\u00e1ch \u0111\u1ec1u hai c\u1ea1nh c\u1ee7a g\u00f3c th\u00ec n\u1eb1m tr\u00ean tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c \u0111\u00f3. ","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Ch\u1ee9ng minh $AD = AE$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D18.png' \/><\/center> <br\/> $\\blacktriangleright$ X\u00e9t hai tam gi\u00e1c vu\u00f4ng $ABD$ v\u00e0 $ACE$ c\u00f3: <br\/> $\\begin{cases} \\widehat{A_{1}} = \\widehat{A_{2}} (gt) \\\\ AB = AC (\\triangle{ABC} \\hspace{0,2cm} \\text{c\u00e2n}) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{ABD} = \\triangle{ACE}$ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow$ $AD = AE$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow$ $KA$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{DKE}$ <br\/> <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span>","column":2}]}],"id_ques":1939},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Cho tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$. G\u1ecdi $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$. T\u1eeb $I$ v\u1ebd $IH \\perp AB; IK \\perp AC$. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang. ","select":["A. $\\triangle{IBH} = \\triangle{ICK}$ ","B. $AH = AK$ ","C. $AI$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{BAC}$ ","D. C\u1ea3 $3$ \u0111\u00e1p \u00e1n tr\u00ean \u0111\u1ec1u \u0111\u00fang"],"hint":"","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv1/img\/H7C3B21_D19.png' \/><\/center> <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{IBH}$ v\u00e0 $\\triangle{ICK}$ c\u00f3: <br\/> $\\begin{cases} \\widehat{BHI} = \\widehat{CKI} = 90^{o} \\\\ IB = IC (gt) \\\\ \\widehat{B} = \\widehat{C} (\\triangle{ABC} \\hspace{0,2cm} \\text{c\u00e2n}) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{BHI} = \\triangle{CKI}$ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) <br\/> $\\blacktriangleright$ $\\Rightarrow$ $IH = IK$ $\\Rightarrow$ $AI$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{BAC}$ (t\u00ednh ch\u1ea5t tia ph\u00e2n gi\u00e1c) <br\/> $\\blacktriangleright$ $\\Rightarrow$ $BH = CK$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> M\u1eb7t kh\u00e1c ta c\u00f3: $AB = AC$ ($\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $AH = AK$ <br\/> <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span>","column":2}]}],"id_ques":1940}],"lesson":{"save":0,"level":1}}