{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["150"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" Tr\u00ean \u0111\u01b0\u1eddng tr\u00f2n $\\left( O;\\,R \\right),$ v\u1ebd c\u00e1c d\u00e2y $AB, AC$ sao cho tia $AO$ n\u1eb1m gi\u1eefa hai tia $AB$ v\u00e0 $AC$, bi\u1ebft $AB = R,$ $\\text{s\u0111}\\overset\\frown{AC}={{90}^{o}}$. S\u1ed1 \u0111o cung nh\u1ecf $BC$ l\u00e0 _input_ $^o$","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai13/lv3/img\/h931_K1.png' \/><\/center> <br\/> X\u00e9t $\\Delta OAB$ c\u00f3: <br\/> $OA = OB = AB = R$ <br\/> $\\Rightarrow \\Delta OAB$ \u0111\u1ec1u (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c \u0111\u1ec1u) <br\/> $\\Rightarrow \\widehat{AOB}={{60}^{o}}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c \u0111\u1ec1u) <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{AB}={{60}^{o}}$ (\u0111\u1ecbnh ngh\u0129a s\u1ed1 \u0111o cung) <br\/> Ta c\u00f3: $\\text{s\u0111}\\overset\\frown{BC}=\\text{s\u0111}\\overset\\frown{AB}+\\text{s\u0111}\\overset\\frown{AC}$ (do $A$ n\u1eb1m gi\u1eefa $B$ v\u00e0 $C$) <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{BC}={{60}^{o}}+{{90}^{o}}={{150}^{o}}$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $150$ <\/span><\/span> "}]}],"id_ques":1431},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["200"]]],"list":[{"point":10,"width":50,"type_input":"","input_hint":["frac"],"ques":" <span class='basic_left'> Cho h\u00ecnh v\u1ebd, <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai13/lv3/img\/h931_K2.png' \/><\/center> <br\/> Bi\u1ebft $\\widehat{BAD}={{33}^{o}};\\widehat{CAD}={{67}^{o}}$ v\u00e0 $AD$ l\u00e0 \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n $(O)$. T\u00ednh s\u1ed1 \u0111o c\u1ee7a $\\overset\\frown{BDC}$. <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $\\text{s\u0111}\\overset\\frown{BDC}=$ _input_$^o$ ","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai13/lv3/img\/h931_K2.1.png' \/><\/center> <br\/>X\u00e9t $\\Delta OAB$ c\u00f3: <br\/> $OA = OB = R$ <br\/> $\\Rightarrow \\Delta OAB$ c\u00e2n (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow \\widehat{BAO}=\\widehat{ABO}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> M\u00e0 $\\,\\widehat{BOD}=\\widehat{BAO}+\\widehat{ABO}$ (\u0111\u1ecbnh l\u00ed g\u00f3c ngo\u00e0i t\u1ea1i m\u1ed9t \u0111\u1ec9nh c\u1ee7a tam gi\u00e1c) <br\/> $\\Rightarrow \\widehat{BOD}=2\\widehat{BAO}={{2.33}^{o}}={{66}^{o}}$ <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{BD}={{66}^{o}}$ (\u0111\u1ecbnh ngh\u0129a s\u1ed1 \u0111o cung) <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1, ta t\u00ednh \u0111\u01b0\u1ee3c: $\\text{s\u0111}\\overset\\frown{DC}={{134}^{o}}$ <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{BDC} = \\text{s\u0111}\\overset\\frown{BD}+\\text{s\u0111}\\overset\\frown{DC}$ ($D$ n\u1eb1m gi\u1eefa $B$ v\u00e0 $C$) <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{BDC} = {{66}^{o}}+{{134}^{o}} = {{200}^{o}}$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $200$<\/span> <\/span>"}]}],"id_ques":1432},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["280"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" <span class='basic_left'>Cho tam gi\u00e1c $ABC$, c\u00f3 $\\widehat{A}={{70}^{o}},\\widehat{B}=35^o.$ V\u1ebd \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m $B$, b\u00e1n k\u00ednh $BA$ v\u00e0 \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m $C$, b\u00e1n k\u00ednh $CA$, ch\u00fang c\u1eaft nhau \u1edf $D$ (kh\u00e1c $A$). T\u00ednh t\u1ed5ng s\u1ed1 \u0111o cung nh\u1ecf $AD$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m $B$ v\u00e0 s\u1ed1 \u0111o cung l\u1edbn $AD$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m $C$. <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> T\u1ed5ng s\u1ed1 \u0111o c\u1ee7a hai cung l\u00e0 _input_$^o$","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai13/lv3/img\/h931_K3.png' \/><\/center> <br\/> G\u1ecdi $\\overset\\frown{ApD}$ l\u00e0 cung nh\u1ecf c\u1ee7a $(C; CA)$ v\u00e0 $\\overset\\frown{AqD}$ l\u00e0 cung l\u1edbn c\u1ee7a $(C; CA)$ <br\/> X\u00e9t $\\Delta BAC$ c\u00f3: <br\/> $\\widehat{BAC}+\\widehat{ABC}+\\widehat{ACB}={{180}^{o}}$ (t\u1ed5ng ba g\u00f3c c\u1ee7a tam gi\u00e1c) <br\/> $\\Rightarrow \\widehat{ACB}={{180}^{o}}-\\widehat{BAC}-\\widehat{ABC}$ <br\/> $={{180}^{o}}-{{70}^{o}}-{{35}^{o}}={{75}^{o}}$ <br\/> X\u00e9t $\\Delta BAC$v\u00e0 $\\Delta BDC$ c\u00f3: <br\/> $AB = BD$ (c\u00f9ng l\u00e0 b\u00e1n k\u00ednh c\u1ee7a $(B)$) <br\/> $AC = DC$ (c\u00f9ng l\u00e0 b\u00e1n k\u00ednh c\u1ee7a $(C)$) <br\/> $BC$ chung <br\/> $\\Rightarrow \\Delta BAC=\\Delta BDC\\,(c.c.c)$ <br\/> $\\Rightarrow \\left\\{ \\begin{align} & \\widehat{ABC}=\\widehat{DBC}={{35}^{o}} \\\\ & \\widehat{ACB}=\\widehat{DCB}={{75}^{o}} \\\\ \\end{align} \\right.$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow \\left\\{ \\begin{align} & \\widehat{ABD}=\\widehat{ABC}+\\widehat{DBC}={{2.35}^{o}}={{70}^{o}} \\\\ & \\widehat{ACD}=\\widehat{ACB}+\\widehat{DCB}={{2.75}^{o}}={{150}^{o}} \\\\ \\end{align} \\right.$ <br\/> $\\Rightarrow \\left\\{ \\begin{align} & \\text{s\u0111}\\overset\\frown{ACD}={{70}^{o}} \\\\ & \\text{s\u0111}\\overset\\frown{ApD}={{150}^{o}} \\\\ \\end{align} \\right.$ (\u0111\u1ecbnh ngh\u0129a s\u1ed1 \u0111o cung) <br\/> Ta c\u00f3: $\\text{s\u0111}\\overset\\frown{AqD}={{360}^{o}}-{{150}^{o}}={{210}^{o}}$ <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{ACD}+\\text{s\u0111}\\overset\\frown{AqD}={{70}^{o}}+{{210}^{o}}={{280}^{o}}$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $280$ <\/span> <\/span>"}]}],"id_ques":1433},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u (>; <; =) th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["<"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" <span class='basic_left'> Cho hai \u0111\u01b0\u1eddng tr\u00f2n $\\left( O;\\,R \\right)\\,$v\u00e0 $\\,\\left( O';\\,R' \\right)$ c\u1eaft nhau t\u1ea1i hai \u0111i\u1ec3m $A$ v\u00e0 $B$ ($R < R\u2019$). K\u1ebb hai \u0111\u01b0\u1eddng k\u00ednh $BOC$ v\u00e0 $BO\u2019D$. So s\u00e1nh s\u1ed1 \u0111o hai cung nh\u1ecf $AC$ v\u00e0 $AD$. <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $\\text{s\u0111}\\overset\\frown{AC}$ _input_ $\\text{s\u0111}\\overset\\frown{AD}$","hint":"S\u1eed d\u1ee5ng quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong 1 tam gi\u00e1c \u0111\u1ec3 so s\u00e1nh hai g\u00f3c \u1edf t\u00e2m r\u1ed3i k\u1ebft lu\u1eadn","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai13/lv3/img\/h931_K4.png' \/><\/center> <br\/> D\u1ec5 d\u00e0ng ch\u1ee9ng minh $C; A; D$ th\u1eb3ng h\u00e0ng <br\/> Ta c\u00f3: $R < R\u2019$ <br\/> $\\Rightarrow BC=2R < 2R'=BD$ <br\/> X\u00e9t $\\Delta BCD$ c\u00f3: <br\/> $BC < BD \\Rightarrow \\widehat{BDC}<\\widehat{BCD}$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong tam gi\u00e1c) (1) <br\/> X\u00e9t $\\Delta AOC$ c\u00f3: <br\/> $OA=OC=R$ <br\/> $\\Rightarrow \\Delta AOC$ c\u00e2n tai $O$ (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow \\widehat{CAO}=\\widehat{ACO}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> M\u00e0 $\\widehat{ACO}+\\widehat{CAO}+\\widehat{COA}={{180}^{o}}$ (t\u1ed5ng ba g\u00f3c c\u1ee7a tam gi\u00e1c) <br\/> $\\Rightarrow \\widehat{AOC} = 180^o - \\widehat{OAC} - \\widehat{OCA}$ <br\/> $ = 180^o - 2\\widehat{OAC} = 180^o - 2\\widehat{BCD}$ (2) <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1: $\\widehat{DO'A}= 180^o - 2\\widehat{O'DA} = 180^o - 2\\widehat{BDC}$(3) <br\/> T\u1eeb (1), (2) v\u00e0 (3), suy ra $\\widehat{COA} < \\widehat{DO'A}$ <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{AC}<\\text{s\u0111}\\overset\\frown{AD}$ (\u0111\u1ecbnh ngh\u0129a s\u1ed1 \u0111o cung) <br\/> <span class='basic_pink'> V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0 $<$ <\/span><\/span> "}]}],"id_ques":1434},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u (>; <; =) th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["="]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $\\left( O;R \\right)\\,$v\u00e0 d\u00e2y $AB$ kh\u00f4ng \u0111i qua $O$. Tr\u00ean d\u00e2y $AB$ l\u1ea5y c\u00e1c \u0111i\u1ec3m $M,\\,N$ sao cho $AM = MN = NB$. Tia $OM, \\, ON$ c\u1eaft $(O)$ l\u1ea7n l\u01b0\u1ee3t t\u1ea1i $C$ v\u00e0 $D$. H\u00e3y so s\u00e1nh s\u1ed1 \u0111o hai cung nh\u1ecf $\\overset\\frown{AC}$ v\u00e0 $\\overset\\frown{BD}.$ <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> s\u0111$\\overset\\frown{AC}$ _input_ s\u0111$\\overset\\frown{BD}$ ","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai13/lv3/img\/h931_K5.png' \/><\/center> <br\/>X\u00e9t $\\Delta OAB$ c\u00f3: <br\/> $OA = OB = R$ <br\/> $\\Rightarrow \\Delta OAB$ c\u00e2n (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow \\widehat{BAO}=\\widehat{ABO}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> X\u00e9t $\\Delta AOM$ v\u00e0 $\\,\\Delta BON$ c\u00f3: <br\/> $\\left\\{ \\begin{align} & OA=OB=R \\\\ & \\widehat{OAM}=\\widehat{OBN}(\\text{ch\u1ee9ng minh tr\u00ean}) \\\\ & AM=BN\\left( \\text{gi\u1ea3 thi\u1ebft} \\right) \\\\ \\end{align} \\right.$ <br\/> $\\Rightarrow \\Delta AOM=\\Delta BON\\left( c.g.c \\right)$ <br\/> $\\Rightarrow \\widehat{AOM}=\\widehat{BON}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{AC}=\\text{s\u0111}\\overset\\frown{BD}$ (\u0111\u1ecbnh ngh\u0129a s\u1ed1 \u0111o cung) <br\/> <span class='basic_pink'> V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0 $=$<\/span> <\/span>"}]}],"id_ques":1435},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u (>; <; =) th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["<"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $\\left( O;R \\right)\\,$v\u00e0 d\u00e2y $AB$ kh\u00f4ng \u0111i qua $O$. Tr\u00ean d\u00e2y $AB$ l\u1ea5y c\u00e1c \u0111i\u1ec3m $M,\\,N$ sao cho $AM = MN = NB$. Tia $OM,\\, ON$ c\u1eaft $(O)$ l\u1ea7n l\u01b0\u1ee3t t\u1ea1i $C$ v\u00e0 $D$. H\u00e3y so s\u00e1nh s\u1ed1 \u0111o hai cung nh\u1ecf $\\overset\\frown{BD}$ v\u00e0 $\\overset\\frown{CD}.$ <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> s\u0111$\\overset\\frown{BD}$ _input_ s\u0111$\\overset\\frown{CD}$ ","hint":"G\u1ecdi $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $OB$. S\u1eed d\u1ee5ng t\u00ednh ch\u1ea5t b\u1eafc c\u1ea7u v\u00e0 quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong tam gi\u00e1c $ONI$ \u0111\u1ec3 so s\u00e1nh hai g\u00f3c \u1edf t\u00e2m th\u00f4ng qua so s\u00e1nh v\u1edbi g\u00f3c $ONI.$","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai13/lv3/img\/h931_K6.png' \/><\/center> G\u1ecdi $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $OB$ <br\/> X\u00e9t $\\Delta BMO$ c\u00f3: <br\/> $MN = NB$ (gi\u1ea3 thi\u1ebft) <br\/> $OI = IB$ ($I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $OB$) <br\/> $\\Rightarrow NI$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a $\\Delta BMO$ <br\/> $\\Rightarrow NI\/\/MO$ (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng trung b\u00ecnh) <br\/> $\\Rightarrow \\widehat{INO}=\\widehat{MON}$ (hai g\u00f3c sole trong) (1) <br\/> Ta c\u00f3: $OM < OB\\Rightarrow IN=\\dfrac{OM}{2}<\\dfrac{OB}{2}=IO$ <br\/> $\\Rightarrow \\widehat{NOI}<\\widehat{ONI}$ (quan h\u1ec7 gi\u1eefa c\u1ea1nh v\u00e0 g\u00f3c trong $\\Delta NIO$) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\widehat{NOI}<\\widehat{MON}$ <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{BD}<\\text{s\u0111}\\overset\\frown{CD}$ (\u0111\u1ecbnh ngh\u0129a s\u1ed1 \u0111o cung) <br\/> <span class='basic_pink'> V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0 $<$ <\/span> <\/span>"}]}],"id_ques":1436},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["90"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" <span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $\\left( O;R \\right)\\,$\u0111\u01b0\u1eddng k\u00ednh $AB$, l\u1ea5y c\u00e1c \u0111i\u1ec3m $M, \\, N$ thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n $(O)$ sao cho $AM=BN=R\\,\\left( M\\in \\overset\\frown{AN} \\right).$ Qua $O$, k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi $AN$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n $(O)$ t\u1ea1i hai \u0111i\u1ec3m $C$ v\u00e0 $D$ $\\left( D\\in \\overset\\frown{\\,BN\\,} \\right).$ S\u1ed1 \u0111o cung nh\u1ecf $CM$ l\u00e0 _input_$^o$ ","hint":"T\u00ednh s\u1ed1 \u0111o g\u00f3c $COM$ r\u1ed3i suy ra s\u1ed1 \u0111o cung nh\u1ecf $CM$","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai13/lv3/img\/h931_K7.png' \/><\/center> <br\/> Ta c\u00f3: $OA=OM=AM=R$ <br\/> $\\Rightarrow \\Delta OAM$ \u0111\u1ec1u (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c \u0111\u1ec1u) <br\/> $\\Rightarrow \\widehat{AOM}={{60}^{o}}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c \u0111\u1ec1u) <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{AM}={{60}^{o}}$ (\u0111\u1ecbnh ngh\u0129a s\u1ed1 \u0111o cung) <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta \u0111\u01b0\u1ee3c: $\\text{s\u0111}\\overset\\frown{BN}={{60}^{o}}$ <br\/> M\u00e0 $\\widehat{AOM}+\\widehat{MON}+\\widehat{NOB}=\\widehat{AOB}={{180}^{o}}$ <br\/> $\\Rightarrow \\widehat{MON}={{180}^{o}}-\\widehat{AOM}-\\widehat{NOB}={{60}^{o}}$ <br\/> X\u00e9t $\\Delta MON$ c\u00f3: <br\/> $OM=ON=R$ <br\/> $\\widehat{MON}={{60}^{o}}$ <br\/> $\\Rightarrow \\Delta MON$ \u0111\u1ec1u (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <br\/> $\\Rightarrow MN=R$ (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c \u0111\u1ec1u) <br\/> X\u00e9t t\u1ee9 gi\u00e1c $AMNO$ c\u00f3: <br\/> $OA = AM = MN = NO = R$ <br\/> $\\Rightarrow AMNO$ l\u00e0 h\u00ecnh thoi (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <br\/> $\\Rightarrow AN\\bot MO$ (t\u00ednh ch\u1ea5t h\u00ecnh thoi) <br\/> M\u00e0 $AN\/\/CD$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow MO\\bot CD$ (t\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song) <br\/> $\\Rightarrow \\widehat{MOC}={{90}^{o}}$$\\Rightarrow \\text{s\u0111}\\overset\\frown{MC}={{90}^{o}}$ (\u0111\u1ecbnh ngh\u0129a s\u1ed1 \u0111o cung) <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $90$ <\/span><\/span> "}]}],"id_ques":1437},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u (>; <; =) th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["AB","BA"],["AC","CA"],["BC","CB"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<span class='basic_left'> Cho tam gi\u00e1c nh\u1ecdn $ABC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O; 1)$ c\u00f3 c\u00e1c c\u1ea1nh $AB=\\sqrt{2};AC=\\sqrt{3}.$ H\u00e3y so s\u00e1nh s\u1ed1 \u0111o c\u00e1c cung nh\u1ecf $AB; BC$ v\u00e0 $AC$. <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> s\u0111$\\overset\\frown{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}} < $ s\u0111$\\overset\\frown{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}} < $ s\u0111$\\overset\\frown{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ ","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai13/lv3/img\/h931_K8.png' \/><\/center> <br\/>G\u1ecdi $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AC$ <br\/> $\\Rightarrow \\left\\{ \\begin{align} & OH\\bot AC \\\\ & AH=HC=\\dfrac{\\sqrt{3}}{2} \\\\ \\end{align} \\right.$ (\u0111\u1ecbnh l\u00ed \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/> X\u00e9t $\\Delta AHO$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: <br\/> $\\sin\\widehat{AOH}=\\dfrac{AH}{AO}=\\dfrac{\\dfrac{\\sqrt{3}}{2}}{1}=\\dfrac{\\sqrt{3}}{2}$ <br\/> $\\Rightarrow \\widehat{AOH}={{60}^{o}}$ <br\/> X\u00e9t $\\Delta AOC$ c\u00f3: <br\/> $OA = OC = R$ <br\/> $\\Rightarrow \\Delta AOC$ c\u00e2n t\u1ea1i $O$ <br\/> M\u00e0 $OH\\bot AC$ (ch\u1ee9ng minh tr\u00ean) <br\/> $\\Rightarrow OH$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{AOC}$ <br\/> $\\Rightarrow \\widehat{AOC}=2\\widehat{AOH}={{2.60}^{o}}={{120}^{o}}$ <br\/> Ta c\u00f3: $O{{A}^{2}}+O{{B}^{2}}=1+1=2=A{{B}^{2}}$ <br\/> $\\Rightarrow \\Delta OAB$ vu\u00f4ng t\u1ea1i $O$ (\u0111\u1ecbnh l\u00ed Pitago \u0111\u1ea3o) <br\/> $\\Rightarrow \\widehat{AOB}={{90}^{o}}$ <br\/> L\u1ea1i c\u00f3 $\\widehat{AOB}+\\widehat{BOC}+\\widehat{AOC}={{360}^{o}}$ <br\/> $\\Rightarrow \\widehat{BOC}={{360}^{o}}-\\widehat{AOB}-\\widehat{AOC}$ <br\/> $\\hspace{1,5cm} ={{360}^{o}}-{{90}^{o}}-{{120}^{o}}={{150}^{o}}$ <br\/> $\\Rightarrow \\widehat{AOB}<\\widehat{AOC}<\\widehat{BOC}$ <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{AB}<\\text{s\u0111}\\overset\\frown{AC}<\\text{s\u0111}\\overset\\frown{BC}$ (\u0111\u1ecbnh ngh\u0129a s\u1ed1 \u0111o cung) <br\/> <span class='basic_pink'> V\u1eady c\u00e1c \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0 $AB; AC$ v\u00e0 $BC$ <\/span> <\/span>"}]}],"id_ques":1438},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["90"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<span class='basic_left'> T\u1eeb m\u1ed9t \u0111i\u1ec3m $A$ tr\u00ean \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$ ta \u0111\u1eb7t li\u00ean ti\u1ebfp c\u00e1c cung $\\overset\\frown{AB};\\,\\overset\\frown{BC};\\,\\overset\\frown{CD}$ l\u1ea7n l\u01b0\u1ee3t c\u00f3 c\u00e1c d\u00e2y cung b\u1eb1ng $R;\\,R\\sqrt{2};\\,R\\sqrt{3}.$ <br\/> <b> C\u00e2u 1: <\/b> S\u1ed1 \u0111o cung nh\u1ecf $ AD$ l\u00e0 _input_ $^o$","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai13/lv3/img\/h931_K9.png' \/><\/center> G\u1ecdi $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $CD$ <br\/> X\u00e9t $\\Delta AOB$ c\u00f3: <br\/> $OA = OB = AB = R$ <br\/> $\\Rightarrow \\Delta AOB$ \u0111\u1ec1u (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c \u0111\u1ec1u) <br\/> $\\Rightarrow \\widehat{AOB}={{60}^{o}}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c \u0111\u1ec1u) <br\/> X\u00e9t $\\Delta BOC$ c\u00f3: <br\/> $O{{B}^{2}}+O{{C}^{2}}={{R}^{2}}+{{R}^{2}}=2{{R}^{2}}=B{{C}^{2}}$ <br\/> $\\Rightarrow \\Delta BOC$ vu\u00f4ng t\u1ea1i $O$ (\u0111\u1ecbnh l\u00ed Pitago \u0111\u1ea3o) <br\/> $\\Rightarrow \\widehat{BOC}={{90}^{o}}$ <br\/> Do $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $CD$ (c\u00e1ch d\u1ef1ng) <br\/> $\\Rightarrow \\left\\{ \\begin{align} & OH\\bot CD \\\\ & CH=HD=\\dfrac{R\\sqrt{3}}{2} \\\\ \\end{align} \\right.$ (\u0111\u1ecbnh l\u00ed \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/> X\u00e9t $\\Delta OHC$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: $\\sin \\widehat{HOC}=\\dfrac{HC}{OC}=\\dfrac{\\dfrac{R\\sqrt{3}}{2}}{R}=\\dfrac{\\sqrt{3}}{2}$ <br\/> $\\Rightarrow \\widehat{HOC}={{60}^{o}}$ <br\/> Ta c\u00f3 $OC=OD$ <br\/> $\\Rightarrow \\Delta COD$ c\u00e2n t\u1ea1i $O$ <br\/> $OH\\bot CD \\Rightarrow OH$ l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{COD}$ <br\/> $\\Rightarrow \\widehat{COD}=2\\widehat{HOC}={{2.60}^{o}}={{120}^{o}}$ <br\/> L\u1ea1i c\u00f3: $\\widehat{AOB}+\\widehat{BOC}+\\widehat{COD}+\\widehat{DOA}={{360}^{o}}$ <br\/> $\\Rightarrow \\widehat{DOA}=360-\\widehat{AOB}-\\widehat{BOC}-\\widehat{COD}$ <br\/> $={{360}^{o}}-{{60}^{o}}-{{90}^{o}}-{{120}^{o}}={{90}^{o}}$ <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{AD}={{90}^{o}}$ (\u0111\u1ecbnh ngh\u0129a s\u1ed1 \u0111o cung) <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $90$ <\/span> <\/span>"}]}],"id_ques":1439},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<span class='basic_left'> T\u1eeb m\u1ed9t \u0111i\u1ec3m $A$ tr\u00ean \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$ ta \u0111\u1eb7t li\u00ean ti\u1ebfp c\u00e1c cung $\\overset\\frown{AB};\\,\\overset\\frown{BC};\\,\\overset\\frown{CD}$ l\u1ea7n l\u01b0\u1ee3t c\u00f3 c\u00e1c d\u00e2y cung b\u1eb1ng $R;\\,R\\sqrt{2};\\,R\\sqrt{3}.$ <br\/> <b> C\u00e2u 2: <\/b> T\u1ee9 gi\u00e1c $ABCD$ l\u00e0 h\u00ecnh g\u00ec? <\/span> ","select":["A. H\u00ecnh b\u00ecnh h\u00e0nh ","B. H\u00ecnh thoi ","C. H\u00ecnh ch\u1eef nh\u1eadt","D. H\u00ecnh thang c\u00e2n"],"explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai13/lv3/img\/h931_K9.png' \/><\/center> <br\/> Theo c\u00e2u 1, ta c\u00f3: <br\/> $\\bullet \\Delta AOB$ \u0111\u1ec1u <br\/> $\\Rightarrow \\widehat{ABO}={{60}^{o}}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c \u0111\u1ec1u) <br\/> $\\bullet\\Delta BOC$ vu\u00f4ng t\u1ea1i $O$ <br\/> M\u00e0 $OB = OC =R$ <br\/> $\\Rightarrow \\Delta BOC$ vu\u00f4ng c\u00e2n t\u1ea1i $O$ <br\/> $\\Rightarrow \\widehat{CBO}=\\widehat{BCO}={{45}^{o}}$ <br\/> $\\Rightarrow \\widehat{ABC}=\\widehat{ABO}+\\widehat{CBO}={{60}^{o}}+{{45}^{o}}={{105}^{o}}$ (1) <br\/> X\u00e9t $\\Delta COH$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: <br\/> $\\widehat{COH}+\\widehat{HCO}={{90}^{o}}$ <br\/> $\\Rightarrow \\widehat{HCO}={{90}^{o}}-{{60}^{o}}={{30}^{o}}$ <br\/> $\\Rightarrow \\widehat{BCD}=\\widehat{BCO}+\\widehat{DCO}={{45}^{o}}+{{30}^{o}}={{75}^{o}}$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\widehat{ABC}+\\widehat{BCD}={{105}^{o}}+{{75}^{o}}={{180}^{o}}$ <br\/> $\\Rightarrow AB\/\/CD$ (hai g\u00f3c trong c\u00f9ng ph\u00eda b\u00f9 nhau) <br\/> $\\Rightarrow ABCD$ l\u00e0 h\u00ecnh thang (3) <br\/> X\u00e9t $\\Delta AOD$ c\u00f3: <br\/> $\\widehat{AOD}={{90}^{o}}$ (theo c\u00e2u 1) <br\/> $OA = OD = R$ <br\/> $\\Rightarrow \\Delta OAD$ vu\u00f4ng c\u00e2n t\u1ea1i $O$ <br\/> $\\Rightarrow \\widehat{ADO}={{45}^{o}}$ <br\/> M\u00e0 $\\Delta COD$ c\u00e2n t\u1ea1i $O$ (theo c\u00e2u 1) <br\/> $\\Rightarrow \\widehat{CDO}=\\widehat{DCO}={{30}^{o}}$ <br\/> $\\Rightarrow \\widehat{ADC}=\\widehat{ADO}+\\widehat{ODC}={{45}^{o}}+{{30}^{o}}={{75}^{o}}$ <br\/> $\\widehat{DCB}=\\widehat{DCO}+\\widehat{BCO}={{30}^{o}}+{{45}^{o}}={{75}^{o}}$ <br\/> $\\Rightarrow \\widehat{ADC}=\\widehat{DCB}$ (4) <br\/> T\u1eeb (3) v\u00e0 (4) suy ra $ABCD$ l\u00e0 h\u00ecnh thang c\u00e2n (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D <\/span><\/span>","column":4}]}],"id_ques":1440}],"lesson":{"save":0,"level":3}}