{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'>T\u00ecm \u0111\u1ed9 d\u00e0i c\u1ee7a $x$ trong h\u00ecnh v\u1ebd sau: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai18/lv3/img\/H8C3B5_K102.png' \/><\/center><\/span>","select":["A. $x = 4,5$","B. $x = 3,5$","C. $x = 3,75$","D. $x = 6$"],"hint":"Ch\u1ee9ng minh $EF \/\/ BC$. T\u1eeb \u0111\u00f3 \u00e1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t \u0111\u1ec3 t\u00ecm $x$.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> T\u00ednh t\u1ec9 s\u1ed1 $\\dfrac{EA}{EB}$ v\u00e0 $\\dfrac{FA}{FC}$<br\/><b>B\u01b0\u1edbc 2:<\/b> T\u1eeb t\u1ec9 s\u1ed1 t\u00ecm \u0111\u01b0\u1ee3c \u1edf b\u01b0\u1edbc 1, \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Ta-l\u00e9t \u0111\u1ea3o ch\u1ee9ng minh \u0111\u01b0\u1ee3c $EF$ \/\/ $BC$<br\/><b>B\u01b0\u1edbc 3:<\/b> \u00c1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t t\u00ecm $x$.<\/span> <br\/> <span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai18/lv3/img\/H8C3B5_K102.png' \/><\/center><br\/>Ta c\u00f3:<br\/>$\\left.\\begin{array}{l} \\dfrac{EA}{EB} = \\dfrac{3}{5} \\\\ \\dfrac{FA}{FC} = \\dfrac{4,5}{7,5} = \\dfrac{3}{5} \\end{array} \\right\\}$ $\\Rightarrow \\dfrac{EA}{EB} = \\dfrac{FA}{FC}$ <br\/> $\\Rightarrow EF \/\/ BC$ (\u0111\u1ecbnh l\u00ed Ta-l\u00e9t \u0111\u1ea3o)<br\/>X\u00e9t $\\triangle ABC$ c\u00f3: $EF \/\/ BC$ (ch\u1ee9ng minh tr\u00ean)<br\/> $\\Rightarrow \\dfrac{AE}{AB}=\\dfrac{EF}{BC} $ (h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t)<br\/>$\\Rightarrow \\dfrac{AE}{AE + EB}=\\dfrac{EF}{BC} $<br\/> $\\Rightarrow$ $\\dfrac{3}{3 + 5}=\\dfrac{x}{10} $<br\/> $\\Rightarrow$ $\\dfrac{3}{8}=\\dfrac{x}{10} $<br\/> $\\Rightarrow$ $x = \\dfrac{3.10}{8} = 3,75$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: <span class='basic_pink'>C. $x = 3,75$<\/span>","column":2}]}],"id_ques":1790},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["8"],["10"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$ c\u00f3 $AB = 12cm$, $AC = 18cm$, \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $AD$. Tr\u00ean $AD$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $AE = \\dfrac{2}{3}.AD$. G\u1ecdi $K$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BE$ v\u00e0 $AC$. T\u00ednh \u0111\u1ed9 d\u00e0i $AK$ v\u00e0 $KC$<br\/> <b>\u0110\u00e1p \u00e1n:<\/b> $AK$ = _input_ ($cm$); $KC$ = _input_ ($cm$)<\/span> ","hint":"K\u1ebb $DI \/\/ BK$ ($I \\in AC$)","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai18/lv3/img\/H8C3B5_K101.png' \/><\/center><br\/>$\\blacktriangleright$ $AD$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a $\\triangle{ABC}$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow \\dfrac{BD}{DC} = \\dfrac{AB}{AC}$ (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c)<br\/>$\\Rightarrow \\dfrac{BD}{DC} = \\dfrac{12}{18} = \\dfrac{2}{3}$<br\/>$\\Rightarrow \\dfrac{BD}{BC} = \\dfrac{BD}{DC + BD} = \\dfrac{2}{5}$<br\/>$\\blacktriangleright$ K\u1ebb $DI \/\/ BK$ ($I \\in AC$), ta c\u00f3:<br\/>$\\dfrac{AK}{KI} = \\dfrac{AE}{ED} = \\dfrac{2}{1}$ <b>(1)<\/b><br\/>$\\dfrac{KI}{KC} = \\dfrac{BD}{BC} = \\dfrac{2}{5}$ <b>(2)<\/b><br\/>T\u1eeb (1) v\u00e0 (2), ta c\u00f3:<br\/>$\\dfrac{AK}{KI}.\\dfrac{KI}{KC} = \\dfrac{2}{1}.\\dfrac{2}{5}$ $\\Rightarrow \\dfrac{AK}{KC} = \\dfrac{4}{5} \\Rightarrow \\dfrac{AK}{4} = \\dfrac{KC}{5}$<br\/>$\\blacktriangleright$ \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau, ta c\u00f3:<br\/> $\\dfrac{AK}{4} = \\dfrac{KC}{5} = \\dfrac{AK + KC}{4 + 5} = \\dfrac{AC}{9} = \\dfrac{18}{9} = 2$<br\/>$\\Rightarrow AK = 2.4 = 8 \\text{(cm)}; KC = 2.5 = 10 \\text{(cm)}$<br\/>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 <span class='basic_pink'>8; 10<\/span> "}]}],"id_ques":1791},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$ c\u00f3 $BC = a, CA = b, AB = c$. Bi\u1ebft r\u1eb1ng $\\widehat{A} = 2\\widehat{B}$.<br\/><b> H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t<\/b><\/span>","select":["A. $a^2 = b^2 + bc$","B. $a^2 = b^2 + ac$","C. $a^2 = b^2 + ab$","D. $a^2 = b^2 - bc$"],"hint":"K\u1ebb \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $AD$","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai18/lv3/img\/H8C3B5_K103.png' \/><\/center><br\/>$\\blacktriangleright$ K\u1ebb \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $AD$<br\/>$\\Rightarrow \\dfrac{AB}{AC} = \\dfrac{BD}{DC}$ (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c) <br\/>$\\Rightarrow \\dfrac{DC}{AC} = \\dfrac{BD}{AB}$<br\/>$\\blacktriangleright$ \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau, ta c\u00f3:<br\/>$\\dfrac{DC}{AC} = \\dfrac{BD}{AB} = \\dfrac{DC + BD}{AC + AB} = \\dfrac{a}{c + b}$<br\/>$\\Rightarrow CD = \\dfrac{ab}{c + b}$ <br\/> $\\blacktriangleright$ L\u1ea1i c\u00f3: $\\widehat{A} = 2\\widehat{B}$ $\\Rightarrow \\widehat{B} = \\dfrac{\\widehat{A}}{2}$<br\/>M\u00e0 $\\widehat{A_1} = \\widehat{A_2} = \\dfrac{\\widehat{A}}{2}$ ($AD$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c)<br\/>$\\Rightarrow \\widehat{A_1} = \\widehat{A_2} = \\widehat{B}$<br\/>$\\blacktriangleright$ X\u00e9t $\\triangle{BCA}$ v\u00e0 $\\triangle{ACD}$ c\u00f3: <br\/> + $\\widehat{C}$ chung<br\/>+ $\\widehat{B} = \\widehat{A_2}$ (ch\u1ee9ng minh tr\u00ean)<br\/>$\\Rightarrow \\triangle{BCA} \\backsim \\triangle{ACD}$ (g\u00f3c -g\u00f3c)<br\/>$\\Rightarrow \\dfrac{BC}{AC} = \\dfrac{AC}{CD}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7)<br\/>$\\Rightarrow \\dfrac{a}{b} = \\dfrac{b}{\\dfrac{ab}{b + c}} \\Rightarrow \\dfrac{a}{b} = \\dfrac{b(b + c)}{ab}$ $\\Rightarrow a^2 = b^2 + bc$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>A. $a^2 = b^2 + bc$<\/span> <\/span>","column":2}]}],"id_ques":1792},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho $\\triangle{ABC}$ $\\backsim$ $\\triangle{HIK}$ theo t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng $k = \\dfrac{3}{7}$. T\u00ednh chu vi $\\triangle{HIK}$ bi\u1ebft chu vi $\\triangle{ABC}$ b\u1eb1ng $45cm$.<br\/><\/span>","select":[" A. $C_{\\triangle{HIK}} = 75cm$"," B. $C_{\\triangle{HIK}} = 85cm$","C. $C_{\\triangle{HIK}} = 105cm$","D. $C_{\\triangle{HIK}} = 125cm$"],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\triangle{ABC}$ $\\backsim$ $\\triangle{HIK}$ theo t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng $k = \\dfrac{3}{7}$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow \\begin{cases} AB = \\dfrac{3}{7}. HI \\\\ AC = \\dfrac{3}{7}. HK \\\\ BC = \\dfrac{3}{7}. IK \\end{cases}$ <br\/> $\\Rightarrow AB + AC + BC = \\dfrac{3}{7} (HI + HK + IK)$ <br\/>Hay $C_{\\triangle{ABC}} = \\dfrac{3}{7}.C_{\\triangle{HIK}}$ <br\/>$\\Rightarrow C_{\\triangle{HIK}} = \\dfrac{7}{3} C_{\\triangle{ABC}} = \\dfrac{7}{3}.45 = 105 \\text{(cm)}$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>C. $C_{\\triangle{HIK}} = 105cm$<\/span><br\/><span class='basic_left'><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>Hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng c\u00f3 t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng b\u1eb1ng t\u1ec9 s\u1ed1 chu vi.<\/span> <br\/><br\/> ","column":2}]}],"id_ques":1793},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho $\\triangle{ABC}$ v\u00e0 c\u00e1c \u0111\u01b0\u1eddng cao $BD$, $CE$. T\u00ednh $\\widehat{AED}$ bi\u1ebft $\\widehat{ACB} = 45^o$.<br\/><\/span>","select":[" A. $\\widehat{AED} = 45^o$"," B. $\\widehat{AED} = 60^o$","C. $\\widehat{AED} = 90^o$","D. $\\widehat{AED} = 120^o$"],"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 $\\triangle$ vu\u00f4ng $ABD$ $\\backsim$ $\\triangle$ vu\u00f4ng $ACE$<br\/><b>B\u01b0\u1edbc 2:<\/b> Ch\u1ee9ng minh $\\triangle{ADE}$ $\\backsim$ $\\triangle{ABC}$<\/span><br\/> <span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai18/lv3/img\/H8C3B5_K104.png' \/><\/center><br\/> $\\blacktriangleright$ $BD$, $CE$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a $\\triangle{ABC}$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow \\widehat{AEC} = \\widehat{ADB} = 90^o$ <br\/>$\\blacktriangleright$ X\u00e9t $\\triangle$ vu\u00f4ng $ABD$ v\u00e0 $\\triangle$ vu\u00f4ng $ACE$ c\u00f3: <br\/>$\\widehat{AEC} = \\widehat{ADB} = 90^o$ (ch\u1ee9ng minh tr\u00ean)<br\/>$\\widehat{A} $ chung<br\/>$\\Rightarrow$ $\\triangle$ vu\u00f4ng $ABD$ $\\backsim$ $\\triangle$ vu\u00f4ng $ACE$ (g\u00f3c - g\u00f3c)<br\/>$\\Rightarrow \\dfrac{AD}{AE} = \\dfrac{AB}{AC}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7)<br\/>$\\Rightarrow \\dfrac{AD}{AB} = \\dfrac{AE}{AC}$<br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ADE}$ v\u00e0 $\\triangle{ABC}$ c\u00f3: <br\/>$\\widehat{A} $ chung<br\/> $\\dfrac{AD}{AB} = \\dfrac{AE}{AC}$ (ch\u1ee9ng minh tr\u00ean) <br\/>$\\Rightarrow \\triangle{ADE}$ $\\backsim$ $\\triangle{ABC}$ (c\u1ea1nh-g\u00f3c-c\u1ea1nh)<br\/>$\\Rightarrow \\widehat{AED} = \\widehat{ACB}$ (c\u1eb7p g\u00f3c t\u01b0\u01a1ng \u1ee9ng)<br\/> M\u00e0 $\\widehat{ACB} = 45^o$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow \\widehat{AED} = 45^o$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0<span class='basic_pink'> B. $\\widehat{AED} = 45^o$<\/span> <\/span> <br\/><\/span> ","column":2}]}],"id_ques":1794},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho h\u00ecnh v\u1ebd nh\u01b0 sau: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai18/lv3/img\/H8C3B5_K105.png' \/><\/center><br\/>T\u00ednh $BN, MN$.<\/span>","select":["A. $BN = 10$; $MN = 8$","B. $BN = 9$; $MN = 8$","C. $BN = 8$; $MN = 6$","D. $BN = 12$; $MN = 10$"],"hint":"Ch\u1ee9ng minh $MN \/\/ AC$. T\u1eeb \u0111\u00f3 \u00e1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t \u0111\u1ec3 t\u00ednh $BN$ v\u00e0 $MN$.","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai18/lv3/img\/H8C3B5_K105.png' \/><\/center><br\/>Ta c\u00f3:<br\/>$\\left.\\begin{array}{l} MN \\perp AB \\text{(gi\u1ea3 thi\u1ebft)}\\\\ AC \\perp AB \\text{(gi\u1ea3 thi\u1ebft)} \\end{array} \\right\\}$ $\\Rightarrow MN \/\/ AC$ (t\u00ednh ch\u1ea5t t\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song) <br\/>X\u00e9t $\\triangle ABC$ c\u00f3: $MN \/\/ AC$ (ch\u1ee9ng minh tr\u00ean)<br\/> $\\Rightarrow \\dfrac{BM}{BA}=\\dfrac{BN}{BC} = \\dfrac{MN}{AC}$ (h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t)<br\/>$\\Rightarrow \\dfrac{6}{9}=\\dfrac{BN}{15} = \\dfrac{MN}{12}$<br\/> $\\Rightarrow$ $\\begin{cases}BN = \\dfrac{6.15}{9} = 10 \\\\ MN = \\dfrac{6.12}{9} = 8\\end{cases}$ <br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: <span class='basic_pink'>A. $BN = 10$ v\u00e0 $MN = 8$<\/span><\/span> <br\/><\/span> ","column":2}]}],"id_ques":1795},{"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":[["1","4"]],"list":[{"point":10,"img":"","ques":"Cho c\u00e1c c\u1eb7p tam gi\u00e1c c\u00f3 c\u00e1c c\u1ea1nh c\u00f3 \u0111\u1ed9 d\u00e0i sau \u0111\u00e2y<br\/><b> H\u00e3y ch\u1ecdn nh\u1eefng c\u1eb7p tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng.<\/b>","hint":"","column":2,"number_true":2,"select":["A. 1,5cm; 3cm; 4cm v\u00e0 4,5cm; 9cm; 12cm","B. 2cm; 5cm; 6cm v\u00e0 4cm; 12cm; 10cm ","C. 3cm; 4cm; 5cm v\u00e0 6cm; 8cm; 12cm","D. 2cm; 4cm; 5cm v\u00e0 4cm; 8cm; 10cm"],"explain":"\u0110\u00e1p \u00e1n $A$ \u0111\u00fang v\u00ec: $\\dfrac{1,5}{4,5} = \\dfrac{3}{9} = \\dfrac{4}{12}$ <br\/>\u0110\u00e1p \u00e1n $B$ sai v\u00ec: $\\dfrac{2}{4} \\neq \\dfrac{5}{12} \\neq \\dfrac{6}{10}$<br\/>\u0110\u00e1p \u00e1n $C$ sai v\u00ec: $\\dfrac{3}{6} = \\dfrac{4}{8} \\neq \\dfrac{5}{12}$<br\/>\u0110\u00e1p \u00e1n $D$ \u0111\u00fang v\u00ec: $\\dfrac{2}{4} = \\dfrac{4}{8} = \\dfrac{5}{10}$"}]}],"id_ques":1796},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho h\u00ecnh v\u1ebd nh\u01b0 sau: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai18/lv3/img\/H8C3B5_K106.png' \/><\/center><br\/><b>H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t<\/b><\/span>","select":["A. $\\widehat{ABD} = \\widehat{ACB}$","B. $\\widehat{ABD} > \\widehat{ACB}$","C. $\\widehat{ABD} < \\widehat{ACB}$"],"hint":"","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai18/lv3/img\/H8C3B5_K106.png' \/><\/center><br\/>X\u00e9t $\\triangle{ABC}$ v\u00e0 $\\triangle{ADB}$ c\u00f3: <br\/>$\\widehat{A} $ chung<br\/> $\\dfrac{AD}{AB} = \\dfrac{AB}{AC}$ (v\u00ec $\\dfrac{6}{12} = \\dfrac{12}{24}$)<br\/>$\\Rightarrow \\triangle{ABC}$ $\\backsim$ $\\triangle{ADB}$ (c\u1ea1nh-g\u00f3c-c\u1ea1nh)<br\/>$\\Rightarrow \\widehat{ABD} = \\widehat{ACB}$ (c\u1eb7p g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: <span class='basic_pink'>A. $\\widehat{ABD} = \\widehat{ACB}$<\/span><\/span> <br\/><\/span> ","column":3}]}],"id_ques":1797},{"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":[[["="],["="]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'>Cho tam gi\u00e1c nh\u1ecdn $ABC$, c\u00e1c \u0111\u01b0\u1eddng cao $AD, BE, CF$ c\u1eaft nhau t\u1ea1i $H$. So s\u00e1nh $HA.HD$; $HB.HE$ v\u00e0 $HC.HF$<br\/> <b>\u0110\u00e1p \u00e1n:<\/b> $HA.HD$ _input_ $HB.HE$ _input_ $HC.HF$<\/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\u1ee9ng minh $\\triangle{AHF}$ $\\backsim$ $\\triangle{CHD}$<br\/><b>B\u01b0\u1edbc 2:<\/b> Ch\u1ee9ng minh $\\triangle{AHE}$ $\\backsim$ $\\triangle{BHD}$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai18/lv3/img\/H8C3B5_K107.png' \/><\/center><br\/>$\\blacktriangleright$ X\u00e9t $\\triangle$ vu\u00f4ng $AHF$ v\u00e0 $\\triangle$ vu\u00f4ng $CHD$ c\u00f3: <br\/>$\\widehat{AHF} = \\widehat{CHD}$ (c\u1eb7p g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh)<br\/>$\\Rightarrow$ $\\triangle$ vu\u00f4ng $AHF$ $\\backsim$ $\\triangle$ vu\u00f4ng $CHD$ (g\u00f3c - g\u00f3c)<br\/>$\\Rightarrow \\dfrac{AH}{CH} = \\dfrac{HF}{HD}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7)<br\/>$\\Rightarrow HA.HD = HC.HF$ <b>(1)<\/b><br\/>$\\blacktriangleright$ X\u00e9t $\\triangle$ vu\u00f4ng $AHE$ v\u00e0 $\\triangle$ vu\u00f4ng $BHD$ c\u00f3: <br\/>$\\widehat{AHE} = \\widehat{BHD}$ (c\u1eb7p g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh)<br\/>$\\Rightarrow$ $\\triangle$ vu\u00f4ng $AHE$ $\\backsim$ $\\triangle$ vu\u00f4ng $BHD$ (g\u00f3c - g\u00f3c)<br\/>$\\Rightarrow \\dfrac{HA}{HB} = \\dfrac{HE}{HD}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7)<br\/>$\\Rightarrow HA.HD = HB.HE$ <b>(2)<\/b><br\/>T\u1eeb (1) v\u00e0 (2) $\\Rightarrow HA.HD = HB.HE = HC.HF$<br\/>V\u1eady c\u00e1c d\u1ea5u c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 <span class='basic_pink'>\"=; =\"<\/span> "}]}],"id_ques":1798},{"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":" <span class='basic_left'> Cho $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$. Tr\u00ean c\u1ea1nh $BC$ l\u1ea5y \u0111i\u1ec3m $K$, t\u1eeb $K$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi $AC$ c\u1eaft $AB$ t\u1ea1i $I$. Bi\u1ebft $BI = 4cm, AB = 12cm$ v\u00e0 $S_{\\triangle{ABC}} = 54m^2$. T\u00ednh $S_{\\triangle{IBK}}$<\/span>","select":[" A. $S_{\\triangle{IBK}} = 4cm^2$"," B. $S_{\\triangle{IBK}} = 5cm^2$","C. $S_{\\triangle{IBK}} = 6cm^2$","D. $S_{\\triangle{IBK}} = 7cm^2$"],"hint":"Ch\u1ee9ng minh $\\triangle{BIK} \\backsim \\triangle{BAC}$, t\u1eeb \u0111\u00f3 suy ra $\\dfrac{S_{\\triangle{BIK}}}{S_{\\triangle{BAC}}}$","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai18/lv3/img\/H8C3B5_K108.png' \/><\/center><br\/>Ta c\u00f3:<br\/>$IK \/\/ AC$ (gi\u1ea3 thi\u1ebft)<br\/>$AB \\bot AC$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow$ $IK \\bot AB$ (\u0111\u1ecbnh l\u00ed t\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song)<br\/>Hay $\\widehat{BIK} = 90^o$<br\/>X\u00e9t $\\triangle$ vu\u00f4ng $BIK$ v\u00e0 $\\triangle$ vu\u00f4ng $BAC$ c\u00f3:<br\/>$\\widehat{BIK} = \\widehat{BAC}$ (c\u00f9ng $ = 90^o$)<br\/>$\\widehat{B}$ chung<br\/>$\\Rightarrow$ $\\triangle$ vu\u00f4ng $BIK$ $\\backsim$ $\\triangle$ vu\u00f4ng $BAC$ (g\u00f3c - g\u00f3c)<br\/>$\\Rightarrow \\left(\\dfrac{BI}{BA}\\right)^2 = \\dfrac{S_{\\triangle{BIK}}}{S_{\\triangle{BAC}}} $ hay $ \\dfrac{S_{\\triangle{BIK}}}{S_{\\triangle{BAC}}} = \\left(\\dfrac{4}{12}\\right)^2 = \\dfrac{1}{9}$ <br\/>$\\Rightarrow S_{\\triangle{BIK}} = \\dfrac{S_{\\triangle{BAC}}}{9} = \\dfrac{54}{9} = 6 (\\text{cm}^2)$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>C. $S_{\\triangle{IBK}} = 6 (\\text{cm}^2)$<\/span><br\/><span class='basic_left'><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>T\u1ec9 s\u1ed1 di\u1ec7n t\u00edch c\u1ee7a hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng b\u1eb1ng t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng<\/span> <br\/><br\/>","column":2}]}],"id_ques":1799}],"lesson":{"save":0,"level":3}}