{"segment":[{"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'> C\u00f3 bao nhi\u00eau v\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a hai \u0111\u01b0\u1eddng tr\u00f2n? <\/span>","select":["A. $2$","B. $3$","C. $4$","D. $5$"],"explain":" <span class='basic_left'> V\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a hai \u0111\u01b0\u1eddng tr\u00f2n l\u00e0: <br\/> - C\u1eaft nhau t\u1ea1i hai \u0111i\u1ec3m ph\u00e2n bi\u1ec7t <br\/> - Ti\u1ebfp x\u00fac trong <br\/> - Ti\u1ebfp x\u00fac ngo\u00e0i <br\/> - \u1ede ngo\u00e0i nhau <br\/> - \u0110\u1ef1ng nhau <br\/> V\u1eady c\u00f3 5 v\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a hai \u0111\u01b0\u1eddng tr\u00f2n <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span>","column":4}]}],"id_ques":1231},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ \u0111\u1ec1u c\u1ea1nh b\u1eb1ng a. V\u1ebd c\u00e1c \u0111\u01b0\u1eddng tr\u00f2n $(A;AB),\\,(C;AC),\\,(B;BC)$. T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c t\u1ea1o b\u1edfi c\u00e1c giao \u0111i\u1ec3m c\u1ee7a ba \u0111\u01b0\u1eddng tr\u00f2n. <\/span>","select":["A. $a^2$","B. $a^2\\sqrt{2}$","C. $a^2\\sqrt{3}$","D. $2a^2$"],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai10/lv3/img\/h925_K4.png' \/><\/center> G\u1ecdi $A',B',C'$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 giao \u0111i\u1ec3m c\u1ee7a ba \u0111\u01b0\u1eddng tr\u00f2n <br\/> D\u1ec5 th\u1ea5y: $A'B'=B'C'=C'A'=2a$ <br\/> Suy ra $\\Delta A'B'C'$ \u0111\u1ec1u <br\/> $A'C$ l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn n\u00ean c\u0169ng l\u00e0 \u0111\u01b0\u1eddng cao <br\/> \u0110\u01b0\u1eddng cao trong tam gi\u00e1c \u0111\u1ec1u c\u1ea1nh $a$ c\u00f3 \u0111\u1ed9 d\u00e0i b\u1eb1ng $\\dfrac{a\\sqrt{3}}{2}$ <br\/> $\\Rightarrow A'C=\\dfrac{A'B'\\sqrt{3}}{2}=a\\sqrt{3}$<br\/> $S_{A'B'C'}=\\dfrac{1}{2}A'C.B'C'$ <br\/> $=\\dfrac{1}{2}.a\\sqrt{3}.2a $ <br\/> $=a^2\\sqrt{3}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span>","column":4}]}],"id_ques":1232},{"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'> Cho tam gi\u00e1c $MNP$ \u0111\u1ec1u c\u1ea1nh b\u1eb1ng $5cm.$ V\u1ebd c\u00e1c \u0111\u01b0\u1eddng tr\u00f2n $(M;MN),$$\\,(P;MP),$$\\,(N;NP)$. T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c t\u1ea1o b\u1edfi c\u00e1c giao \u0111i\u1ec3m c\u1ee7a ba \u0111\u01b0\u1eddng tr\u00f2n. <\/span>","select":["A. $25\\,cm^2$","B. $50\\,cm^2$","C. $25\\sqrt{2}\\,cm^2$","D. $25\\sqrt{3}\\,cm^2$"],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai10/lv3/img\/h925_K4.1.png' \/><\/center> <br\/> \u00c1p d\u1ee5ng k\u1ebft qu\u1ea3 c\u1ee7a c\u00e2u tr\u00ean<br\/> $S_{MNP}=a^2\\sqrt{3}$$=25\\sqrt{3} (cm^2)$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span>","column":4}]}],"id_ques":1233},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho hai \u0111\u01b0\u1eddng tr\u00f2n $(O)$ v\u00e0 $(O')$ ti\u1ebfp x\u00fac ngo\u00e0i t\u1ea1i $A$. K\u1ebb hai ti\u1ebfp tuy\u1ebfn chung ngo\u00e0i $MN$ v\u00e0 $PQ$ v\u1edbi $M, P$ c\u00f9ng thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n $(O)$; $N, Q$ c\u00f9ng thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n $(O')$. Hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau t\u1ea1i $S$. <br\/> <b> C\u00e2u 1: <\/b> Tam gi\u00e1c $AMN$ l\u00e0 tam gi\u00e1c g\u00ec? <\/span>","select":["A. Tam gi\u00e1c c\u00e2n","B. Tam gi\u00e1c vu\u00f4ng","C. Tam gian vu\u00f4ng c\u00e2n","D. Tam gi\u00e1c \u0111\u1ec1u"],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai10/lv3/img\/h925_K1.png' \/><\/center> Ta c\u00f3 $MN$ l\u00e0 ti\u1ebfp tuy\u1ebfn chung c\u1ee7a $(O)$ v\u00e0 $(O')$ <br\/> $\\Rightarrow OM\\bot SM;O'N\\bot SM $ (t\u00ednh ch\u1ea5t ti\u1ebfp tuy\u1ebfn) <br\/> $\\Rightarrow O'N\/\/OM $ (t\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song) <br\/> $ \\Rightarrow \\widehat{NO'A}+\\widehat{AOM}={{180}^{o}}$ (t\u1ed5ng hai g\u00f3c trong c\u00f9ng ph\u00eda) <br\/>M\u00e0 $OM=OA$ $\\Rightarrow \\Delta OAM$ c\u00e2n t\u1ea1i $O$ (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow\\widehat{{{A}_{1}}}=\\widehat{{{M}_{1}}}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/>M\u00e0 $O'N=O'A\\Rightarrow \\Delta O'AN$ c\u00e2n t\u1ea1i $O'$ (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow\\widehat{{{A}_{2}}}=\\widehat{{{N}_{1}}}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> $\\widehat{NO'A}+\\widehat{AOM}$$+\\widehat{{{A}_{1}}} +\\widehat{{{M}_{1}}}+\\widehat{{{A}_{2}}}+\\widehat{{{N}_{1}}}$$ = 180^o + 180^o={{360}^{o}}$ (t\u1ed5ng s\u1ed1 \u0111o c\u1ee7a hai tam gi\u00e1c) <br\/> $\\Rightarrow 180^o + 2\\left( \\widehat{{{A}_{1}}}+\\widehat{{{A}_{2}}} \\right)={{360}^{o}} $ <br\/> $\\Rightarrow 2\\left( \\widehat{{{A}_{1}}}+\\widehat{{{A}_{2}}} \\right)={{180}^{o}} $ <br\/> $ \\Rightarrow \\widehat{{{A}_{1}}}+\\widehat{{{A}_{2}}}={{90}^{o}} $ <br\/> $\\Rightarrow \\widehat{MAN}={{90}^{o}} $ <br\/> $\\Rightarrow \\Delta AMN$ vu\u00f4ng t\u1ea1i $A$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span>","column":2}]}],"id_ques":1234},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho hai \u0111\u01b0\u1eddng tr\u00f2n $(O)$ v\u00e0 $(O')$ ti\u1ebfp x\u00fac ngo\u00e0i t\u1ea1i $A$. K\u1ebb hai ti\u1ebfp tuy\u1ebfn chung ngo\u00e0i $MN$ v\u00e0 $PQ$ v\u1edbi $M, P$ c\u00f9ng thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n $(O)$; $N, Q$ c\u00f9ng thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n $(O')$. Hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau t\u1ea1i $S$. <br\/> <b> C\u00e2u 2: <\/b> T\u1ee9 gi\u00e1c $MNQP$ l\u00e0 h\u00ecnh g\u00ec? <\/span>","select":["A. H\u00ecnh thang","B. H\u00ecnh thang vu\u00f4ng","C. H\u00ecnh thang c\u00e2n"],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai10/lv3/img\/h925_K1.1.png' \/><\/center> Ta c\u00f3 $SM=SP; SN=SQ$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> $\\Rightarrow MN=PQ$ <br\/> Trong tam gi\u00e1c $MSP$ c\u00f3: <br\/> $ \\dfrac{SN}{SM}=\\dfrac{SQ}{SP}$ <br\/> $\\Rightarrow NQ\/\/MP $ (\u0111\u1ecbnh l\u00ed Ta - let \u0111\u1ea3o) <br\/> $\\Rightarrow MNQP$ l\u00e0 h\u00ecnh thang (1) <br\/>Ta c\u00f3: $SM=SP$ (ch\u1ee9ng minh tr\u00ean) <br\/> $\\Rightarrow \\Delta SMN$ c\u00e2n t\u1ea1i $S$ <br\/> $\\Rightarrow \\widehat{PMS} = \\widehat{MPS}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> Hay $ \\widehat{PMN} = \\widehat{MPQ}$ (2) <br\/> T\u1eeb (1), (2) $\\Rightarrow MNQP$ 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 C.<\/span><\/span>","column":3}]}],"id_ques":1235},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank","correct":[[["40"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<span class='basic_left'>Cho hai \u0111\u01b0\u1eddng tr\u00f2n $(O)$ v\u00e0 $(O')$ ti\u1ebfp x\u00fac ngo\u00e0i t\u1ea1i $A$. K\u1ebb hai ti\u1ebfp tuy\u1ebfn chung ngo\u00e0i $MN$ v\u00e0 $PQ$ v\u1edbi $M, P$ c\u00f9ng thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n $(O)$; $N, Q$ c\u00f9ng thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n $(O')$. Hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau t\u1ea1i $S$. <br\/><b> C\u00e2u 3: <\/b> Cho $OM = 15cm;\\,O'N=3cm;\\, OO'=20cm$. Khi \u0111\u00f3: $SM+SP= $_input_ $(cm)$<\/span>","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai10/lv3/img\/h925_K1.1.png' \/><\/center><span class='basic_left'> Theo c\u00e2u 1: $O'N\/\/OM$$\\Rightarrow \\dfrac{SO'}{SO}=\\dfrac{O'N}{OM}$ (\u0111\u1ecbnh l\u00ed Ta - let) <br\/> $\\Leftrightarrow SO'.OM=SO.O'N $ <br\/> $\\Leftrightarrow SO'.15=(SO'+20).3 $ <br\/> $\\Leftrightarrow 12.SO'=60 $ <br\/> $\\Rightarrow SO'=5(cm)$ <br\/> $\\Rightarrow SO=SO'+OO'=5+20=25 \\,(cm)$ <br\/> X\u00e9t $\\Delta SMO$ vu\u00f4ng t\u1ea1i $ M $ <br\/> $SO^2 = SM^2 + OM^2$ (\u0111\u1ecbnh l\u00ed Pitago) <br\/> $\\Rightarrow SM=\\sqrt{S{{O}^{2}}-O{{M}^{2}}}$ <br\/> $=\\sqrt{{{25}^{2}}-{{15}^{2}}}$ <br\/> $=20 \\,(cm)$ <br\/> $\\Rightarrow SM + SP = 2SM = 40 \\,(cm)$ <br\/> <span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $40.$<\/span><\/span>"}]}],"id_ques":1236},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{5}{2}$","B. $\\dfrac{2}{3}$","C. $\\dfrac{1}{3}$"],"ques":"<span class='basic_left'>Cho ba \u0111i\u1ec3m $I, J, K$ c\u00f9ng n\u1eb1m tr\u00ean m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng sao cho $IJ=10cm,\\,IK=14cm$. V\u1ebd \u0111\u01b0\u1eddng tr\u00f2n $(O)$ \u0111\u01b0\u1eddng k\u00ednh $IJ$ v\u00e0 \u0111\u01b0\u1eddng tr\u00f2n $(O')$ \u0111\u01b0\u1eddng k\u00ednh $JK$. <br\/><b> C\u00e2u 1: <\/b> G\u1ecdi $A$ l\u00e0 \u0111i\u1ec3m n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng tr\u00f2n $(O)$, tia $AJ$ giao v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n $(O')$ t\u1ea1i $A'$. <br\/> Khi \u0111\u00f3, $\\dfrac{AJ}{JA'}= $?<\/span>","hint":"X\u00e9t tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng r\u1ed3i l\u1eadp t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng \u0111\u1ec3 t\u00ecm k\u1ebft qu\u1ea3","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai10/lv3/img\/h925_K2.png' \/><\/center><span class='basic_left'>Ta c\u00f3: $IJ=10cm;IK=14cm$ <br\/> $\\Rightarrow JK = IK - IJ = 14 - 10 = 4\\,(cm)$ <br\/> $OA = OI = OJ$ <br\/> $\\Rightarrow \\Delta{IAJ}$ l\u00e0 tam gi\u00e1c vu\u00f4ng (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Rightarrow \\widehat{IAJ}=90^o$ <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1: $\\widehat{JA'K}={{90}^{o}}$ <br\/> X\u00e9t $\\Delta IAJ$ v\u00e0 $\\Delta KA'J:$ <br\/> $\\widehat{IAJ}=\\widehat{JA'K}={{90}^{o}}$ <br\/> $\\widehat{{{J}_{1}}}=\\widehat{{{J}_{2}}}$ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh) <br\/> $\\Rightarrow \\Delta IAJ\\sim \\Delta KA'J$ (g.g) <br\/> $\\Rightarrow \\dfrac{IJ}{JK}=\\dfrac{AJ}{JA'}=\\dfrac{10}{4}=\\dfrac{5}{2}$<\/span>"}]}],"id_ques":1237},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["JA","AJ"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<span class='basic_left'>Cho ba \u0111i\u1ec3m $I, J, K$ c\u00f9ng n\u1eb1m tr\u00ean m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng sao cho $IJ=10cm,\\,IK=14cm$. V\u1ebd \u0111\u01b0\u1eddng tr\u00f2n $(O)$ \u0111\u01b0\u1eddng k\u00ednh $IJ$ v\u00e0 \u0111\u01b0\u1eddng tr\u00f2n $(O')$ \u0111\u01b0\u1eddng k\u00ednh $JK$. <br\/><b> C\u00e2u 2: <\/b> Qua $J$ k\u1ebb m\u1ed9t c\u00e1t tuy\u1ebfn c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n $(O)$ t\u1ea1i $B$ ($B$ v\u00e0 $A$ thu\u1ed9c hai n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $IJ$), c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n $(O')$ t\u1ea1i $B'$.<br\/> Khi \u0111\u00f3 $JA'.AB=A'B'. $_input_ <\/span>","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai10/lv3/img\/h925_K2.2.png' \/><\/center><span class='basic_left'> Ta c\u00f3: $OJ = \\dfrac{IJ}{2} = \\dfrac{10}{2} = 5\\,(cm)$ <br\/> $O'J= \\dfrac{JK}{2} = \\dfrac{4}{2} = 2\\,(cm)$ <br\/> $\\Rightarrow \\dfrac{JB}{JB'} = \\dfrac{OJ}{O'J}=\\dfrac{5}{2}$ <br\/> D\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c: $\\widehat{OBJ}=\\widehat{BJO}$; $\\widehat{O'JB'}=\\widehat{O'B'J}$ <br\/> $\\widehat{BJO}=\\widehat{B'JO'}$(hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh) <br\/> $\\Rightarrow \\widehat{OBJ}=\\widehat{O'B'J}$ <br\/> $\\Rightarrow O'B'\/\/OB$ (hai g\u00f3c so le trong) <br\/> Theo c\u00e2u 1:$\\dfrac{JA}{JA'}=\\dfrac{5}{2}$ $\\Rightarrow \\dfrac{JA}{JA'}=\\dfrac{JB}{JB'}$ (\u0111\u1ecbnh l\u00ed Ta-l\u00e9t) <br\/> X\u00e9t $\\Delta JAB$ v\u00e0 $ \\Delta JA'B':$ <br\/> $\\dfrac{JA}{JA'}=\\dfrac{JB}{JB'};\\widehat{BJA}=\\widehat{B'JA'}$ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh) <br\/> $\\Rightarrow \\Delta JAB\\sim \\Delta JA'B'(c.g.c)$ <br\/> $\\Rightarrow \\dfrac{JA}{JA'}=\\dfrac{AB}{A'B'}$ <br\/> $\\Rightarrow JA'.AB=JA.A'B'$ <br\/> <span class='basic_pink'>V\u1eady t\u1eeb c\u1ea7n \u0111i\u1ec1n l\u00e0 $JA.$<\/span><\/span>"}]}],"id_ques":1238},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'>Cho ba \u0111i\u1ec3m $I, J, K$ c\u00f9ng n\u1eb1m tr\u00ean m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng sao cho $IJ=10cm,\\,IK=14cm$. V\u1ebd \u0111\u01b0\u1eddng tr\u00f2n $(O)$ \u0111\u01b0\u1eddng k\u00ednh $IJ$ v\u00e0 \u0111\u01b0\u1eddng tr\u00f2n $(O')$ \u0111\u01b0\u1eddng k\u00ednh $JK$. Qua $J$ k\u1ebb m\u1ed9t c\u00e1t tuy\u1ebfn c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n $(O)$ t\u1ea1i $B$ ($B$ v\u00e0 $A$ thu\u1ed9c hai n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $IJ$), c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n $(O')$ t\u1ea1i $B'$. <br\/> <b> C\u00e2u 3: Ch\u1ee9ng minh $\\Delta OAB\\sim \\Delta O'A'B'$. <\/span>","title_trans":" H\u00e3y s\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c ch\u1ee9ng minh \u0111\u00fang","temp":"sequence","correct":[[[3],[4],[1],[5],[2]]],"list":[{"point":10,"image":"https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai10/lv3/img\/h925_K2.1.png","left":["M\u00e0 $\\dfrac{OA}{O'A'}=\\dfrac{OB}{O'B'} =\\dfrac{5}{2}$ ","$\\Rightarrow \\dfrac{AB}{A'B'} = \\dfrac{OB}{O'B'} = \\dfrac{OA}{O'A'}$ ","Theo c\u00e2u 2: $\\Delta JAB\\sim \\Delta JA'B'$","$\\Rightarrow \\Delta OAB\\sim \\Delta O'A'B'(c.c.c)$","$\\Rightarrow \\dfrac{AB}{A'B'}=\\dfrac{JA}{JA'}$$=\\dfrac{5}{2}$ (t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng)"],"top":65,"explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai10/lv3/img\/h925_K2.1.png' \/><\/center><span class='basic_left'> Theo c\u00e2u 2: $\\Delta JAB\\sim \\Delta JA'B'$ <br\/> $\\Rightarrow \\dfrac{AB}{A'B'}=\\dfrac{JA}{JA'}$$=\\dfrac{5}{2}$ (t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng) <br\/> M\u00e0 $\\dfrac{OA}{O'A'}=\\dfrac{OB}{O'B'}$$=\\dfrac{5}{2}$ <br\/> $\\Rightarrow \\dfrac{AB}{A'B'} = \\dfrac{OB}{O'B'} = \\dfrac{OA}{O'A'}$ <br\/> $\\Rightarrow \\Delta OAB\\sim \\Delta O'A'B'(c.c.c)$<\/span>"}]}],"id_ques":1239},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho ba \u0111i\u1ec3m $I, J, K$ c\u00f9ng n\u1eb1m tr\u00ean m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng sao cho $IJ=10cm,\\,IK=14cm$. V\u1ebd \u0111\u01b0\u1eddng tr\u00f2n $(O)$ \u0111\u01b0\u1eddng k\u00ednh $IJ$ v\u00e0 \u0111\u01b0\u1eddng tr\u00f2n $(O')$ \u0111\u01b0\u1eddng k\u00ednh $JK$. Qua $J$ k\u1ebb m\u1ed9t c\u00e1t tuy\u1ebfn c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n $(O)$ t\u1ea1i $B$ ($B$ v\u00e0 $A$ thu\u1ed9c hai n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $IJ$), c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n $(O')$ t\u1ea1i $B'$. <br\/> <b> C\u00e2u 4:<\/b> T\u1ee9 gi\u00e1c $ABA'B'$ l\u00e0 h\u00ecnh g\u00ec? <\/span>","select":["A. H\u00ecnh thang","B. H\u00ecnh vu\u00f4ng","C. H\u00ecnh b\u00ecnh h\u00e0nh","D. H\u00ecnh ch\u1eef nh\u1eadt"],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai10/lv3/img\/h925_K2.3.png' \/><\/center> Theo c\u00e2u 2: $\\Delta JAB\\sim \\Delta JA'B' $ <br\/> $\\Rightarrow \\widehat{JA'B'}=\\widehat{JAB}$ <br\/> $\\Rightarrow AB\/\/A'B'$ (hai g\u00f3c so le trong) <br\/> Suy ra $ABA\u2019B\u2019$ l\u00e0 h\u00ecnh thang <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span>","column":2}]}],"id_ques":1240}],"lesson":{"save":0,"level":3}}