{"segment":[{"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 $ABC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O)$. G\u1ecdi $OD, OE, OF$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 kho\u1ea3ng c\u00e1ch t\u1eeb $O$ \u0111\u1ebfn $AB, AC, BC.$ Bi\u1ebft $OE < OF < OD.$ So s\u00e1nh c\u00e1c g\u00f3c c\u1ee7a tam gi\u00e1c $ABC$ <\/span>","select":["A. $\\widehat{A} < \\widehat{B}<\\widehat{C}$","B. $\\widehat{C} < \\widehat{B} < \\widehat{A}$","C. $\\widehat{C} < \\widehat{A}< \\widehat{B}$","D. $\\widehat{B} < \\widehat{C}<\\widehat{A}$"],"explain":"<span class='basic_left'> $OE < OF < OD$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow AC > BC > AB$ (\u0111\u1ecbnh l\u00ed v\u1ec1 kho\u1ea3ng c\u00e1ch t\u1eeb t\u00e2m \u0111\u1ebfn d\u00e2y) <br\/> $\\Delta ABC$ c\u00f3: $AC > BC > AB$ <br\/> $\\Rightarrow$ $\\widehat{B}>\\widehat{A}>\\widehat{C}$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong m\u1ed9t tam gi\u00e1c)<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span>","column":2}]}],"id_ques":1141},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$ \u0111\u01b0\u1eddng k\u00ednh $AB.$ G\u1ecdi $M$ v\u00e0 $N$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $OA, OB.$ Qua $M$ v\u00e0 $N$ l\u1ea7n l\u01b0\u1ee3t v\u1ebd c\u00e1c d\u00e2y $CD$ v\u00e0 $EF$ song song v\u1edbi nhau ($C$ v\u00e0 $E$ n\u1eb1m tr\u00ean m\u1ed9t n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh $AB$)<br\/><b> C\u00e2u 1: <\/b> T\u1ee9 gi\u00e1c $CDFE$ l\u00e0 h\u00ecnh g\u00ec?<\/span>","select":["A. H\u00ecnh thang","B. H\u00ecnh thang c\u00e2n","C. H\u00ecnh ch\u1eef nh\u1eadt","D. H\u00ecnh vu\u00f4ng"],"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai7/lv3/img\/H922_K10.jpg' \/><\/center><br\/> H\u1ea1 OH $\\bot$ $CD$ t\u1ea1i $H,$ $OH$ c\u1eaft $EF$ t\u1ea1i $K$ <br\/> V\u00ec $CD \/\/ EF\\, (1),$ $OH \\bot CD$ <br\/> $\\Rightarrow OK \\bot EF.$ <br\/> X\u00e9t hai tam gi\u00e1c vu\u00f4ng $MHO$ v\u00e0 $NKO$ c\u00f3: <br\/> $MO = ON =\\dfrac {R}{2}$ ($M$ v\u00e0 $N$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $OA, OB$) <br\/> $\\widehat{HOM}=\\widehat{KON}$ (\u0111\u1ed1i \u0111\u1ec9nh) <br\/> $\\Rightarrow \\Delta MHO =\\Delta NKO$ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow OH = OK$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow CD = EF$ (\u0111\u1ecbnh l\u00ed v\u1ec1 kho\u1ea3ng c\u00e1ch t\u1eeb t\u00e2m \u0111\u1ebfn d\u00e2y) (2)<br\/>T\u1eeb (1), (2) $\\Rightarrow$ T\u1ee9 gi\u00e1c $CDFE$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <br\/> M\u00e0 $OH \\bot CD, OK \\bot EF$ <br\/> $ \\Rightarrow CH = HD; EK = KF$ (quan h\u1ec7 vu\u00f4ng g\u00f3c gi\u1eefa \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/> Suy ra $HK$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a h\u00ecnh b\u00ecnh h\u00e0nh $CDFE$ <br\/>M\u00e0 $\\widehat{ HKE} = 90^o$ $\\Rightarrow \\widehat{CEF}=90^o$ <br\/>$\\Rightarrow CDFE$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span>","column":2}]}],"id_ques":1142},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\sqrt{15}$","B. $2\\sqrt{15}$","C. $3\\sqrt{15}$"],"ques":"<span class='basic_left'>Cho \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$ \u0111\u01b0\u1eddng k\u00ednh $AB.$ G\u1ecdi $M$ v\u00e0 $N$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $OA, OB.$ Qua $M$ v\u00e0 $N$ l\u1ea7n l\u01b0\u1ee3t v\u1ebd c\u00e1c d\u00e2y $CD$ v\u00e0 $EF$ song song v\u1edbi nhau ($C$ v\u00e0 $E$ n\u1eb1m tr\u00ean m\u1ed9t n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh $AB$) <br\/><b> C\u00e2u 2: <\/b> Gi\u1ea3 s\u1eed $CD$ v\u00e0 $EF$ c\u00f9ng t\u1ea1o v\u1edbi $AB$ m\u1ed9t g\u00f3c nh\u1ecdn l\u00e0 $30^o, R= 2 cm.$ Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt $CDFE $ l\u00e0 ?cm$^2$<\/span>","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai7/lv3/img\/H922_K10.jpg' \/><\/center><br\/> $\\Delta HOM$ vu\u00f4ng t\u1ea1i $H$ c\u00f3 $\\widehat{M}=30^o$ <br\/> $\\Rightarrow OH =OM.sin{\\widehat{OMH}} =\\dfrac{1}{2}OM $$=\\dfrac{1}{4}R=\\dfrac{1}{2}\\,(cm)$ <br\/> V\u1eady $HK=CE=2OH=1\\,(cm)$. <br\/> X\u00e9t $\\Delta CEF$ c\u00f3 $\\widehat{E}=90^o$ (theo c\u00e2u 1) <br\/> $\\Rightarrow$ t\u00e2m $O$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp $\\Delta CEF$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $CF$ <br\/> $\\Rightarrow CF$ l\u00e0 \u0111\u01b0\u1eddng k\u00ednh <br\/> $\\Delta CEF $ c\u00f3: $\\widehat{E}=90^o$ <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pitago trong $\\Delta CEF$ ta c\u00f3: <br\/> $CF^2=EF^2+CE^2\\\\ \\Rightarrow EF^2= 4^2 - 1=15\\\\ \\Rightarrow EF =\\sqrt{15}\\,(cm)$ <br\/> Di\u1ec7n th\u00edch h\u00ecnh ch\u1eef nh\u1eadt $CDFE$ l\u00e0 : <br\/>$S_{CDEF}=CE.EF$$=1.\\sqrt{15}=\\sqrt{15}\\,(cm^2)$<\/span>"}]}],"id_ques":1143},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["4"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Cho \u0111\u01b0\u1eddng tr\u00f2n $(O; 13cm)$ v\u00e0 m\u1ed9t \u0111i\u1ec3m $M$ c\u00e1ch $O$ l\u00e0 $5cm.$<br\/>H\u1ecfi c\u00f3 bao nhi\u00eau d\u00e2y c\u00f3 \u0111\u1ed9 d\u00e0i l\u00e0 m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean \u0111i qua $M.$ <br\/><b> \u0110\u00e1p s\u1ed1: <\/b> C\u00f3 t\u1ea5t c\u1ea3 _input_ d\u00e2y cung. ","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai7/lv3/img\/H922_K11.jpg' \/><\/center><span class='basic_left'> V\u1ebd \u0111\u01b0\u1eddng k\u00ednh $AB$ \u0111i qua \u0111i\u1ec3m $M$ <br\/> $\\Rightarrow AB =26\\, (cm)$<br\/> V\u1ebd d\u00e2y $CD$ \u0111i qua $M$ v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi $AB$ t\u1ea1i $M$ <br\/> X\u00e9t $\\Delta CMO$ vu\u00f4ng t\u1ea1i $M$ c\u00f3: <br\/> $CO^2=CM^2+MO^2$ (\u0111\u1ecbnh l\u00ed Pitago) <br\/> $ \\Rightarrow CM=\\sqrt{OC^2-OM^2}=12 \\,(cm) $ <br\/> $\\Rightarrow CD = 2 CM = 24 \\, (cm)$ <br\/> V\u00ec d\u00e2y l\u1edbn nh\u1ea5t \u0111i qua $M$ l\u00e0 $AB (=26\\,cm)$, d\u00e2y nh\u1ecf nh\u1ea5t \u0111i qua $M$ l\u00e0 $CD (=24\\,cm)$ <br\/> N\u00ean c\u00e1c d\u00e2y c\u00f3 \u0111\u1ed9 d\u00e0i l\u00e0 s\u1ed1 t\u1ef1 nhi\u00ean \u0111i qua $M$ s\u1ebd c\u00f3 \u0111\u1ed9 d\u00e0i l\u1ea7n l\u01b0\u1ee3t l\u00e0: $ 24\\, cm;$ v\u00e0 $25\\, cm$ v\u00e0 $26\\, cm$ <br\/> Do t\u00ednh \u0111\u1ed1i x\u1ee9ng c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n qua \u0111\u01b0\u1eddng k\u00ednh $AB$ <br\/> $\\Rightarrow$ C\u00f3 2 d\u00e2y c\u00f3 \u0111\u1ed9 d\u00e0i $25\\, cm;$ v\u00e0 ch\u1ec9 c\u00f3 1 d\u00e2y c\u00f3 \u0111\u1ed9 d\u00e0i $24\\, cm$ v\u00e0 1 d\u00e2y c\u00f3 \u0111\u1ed9 d\u00e0i $26\\, cm$<br\/> V\u1eady t\u1ea5t c\u1ea3 c\u00f3 $4$ d\u00e2y \u0111i qua $M$ c\u00f3 \u0111\u1ed9 d\u00e0i l\u00e0 m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean<br\/><span class='basic_pink'>K\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $4.$ <\/span><\/span>"}]}],"id_ques":1144},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O)$ . G\u1ecdi $OD, OE, OF$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 kho\u1ea3ng c\u00e1ch t\u1eeb $O$ \u0111\u1ebfn $AB, AC, BC.$ Bi\u1ebft $\\widehat{A}=80^o, \\widehat{B}=50^o$. So s\u00e1nh $OD, OE, OF$ <\/span>","select":["A. $OD > OE > OF$","B. $OD = OE > OF$","C. $OD = OE < OF$","D. $OF > OE > OD$"],"explain":"<span class='basic_left'>X\u00e9t $\\Delta ABC$ c\u00f3: <br\/> $\\widehat{A}+\\widehat{B}+\\widehat{C}=180^o$ (\u0111\u1ecbnh l\u00ed t\u1ed5ng ba g\u00f3c trong m\u1ed9t tam gi\u00e1c) <br\/> $\\Rightarrow 80^o + 50^o + \\widehat{C} = 180^o$ (v\u00ec $\\widehat{A}=80^o, \\widehat{B}=50^o$) <br\/> $ \\Rightarrow \\widehat{C}=50^o$ <br\/> Suy ra $\\widehat{A} > \\widehat{C} = \\widehat{B} $ <br\/> $\\Rightarrow BC > AB = AC $ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong 1 tam gi\u00e1c) <br\/> Do \u0111\u00f3 $OF < OD = OE$ (\u0111\u1ecbnh l\u00ed v\u1ec1 kho\u1ea3ng c\u00e1ch t\u1eeb t\u00e2m \u0111\u1ebfn d\u00e2y) <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span>","column":2}]}],"id_ques":1145},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\sqrt{10}$","B. $2\\sqrt{10}$","C. $3\\sqrt{10}$"],"ques":"<span class='basic_left'>Cho \u0111\u01b0\u1eddng tr\u00f2n $(O; 5)$ v\u00e0 hai b\u00e1n k\u00ednh $OA$ v\u00e0 $OB$ vu\u00f4ng g\u00f3c v\u1edbi nhau. Tr\u00ean c\u00e1c b\u00e1n k\u00ednh $OA, OB$ l\u1ea7n l\u01b0\u1ee3t l\u1ea5y c\u00e1c \u0111i\u1ec3m $M$ v\u00e0 $N$ sao cho $OM = ON = \\sqrt{5}$. V\u1ebd d\u00e2y $CD$ \u0111i qua $M$ v\u00e0 $N$ ($M$ n\u1eb1m gi\u1eefa $C$ v\u00e0 $N$). <br\/><b> C\u00e2u 1: <\/b> \u0110\u1ed9 d\u00e0i $MN =$ ?<\/span>","hint":"\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pitago","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai7/lv3/img\/H922_K7.jpg' \/><\/center><br\/> <span class='basic_left'> V\u1ebd $OH \\bot CD$ t\u1ea1i $H.$ \u0110\u1eb7t $OH = x.$ <br\/> Ta c\u00f3: $\\widehat{AOB}=90^o, OM = ON$ (gi\u1ea3 thi\u1ebft)<br\/> $ \\Rightarrow$ $\\Delta MON$ vu\u00f4ng c\u00e2n <br\/> Suy ra $OH$ v\u1eeba l\u00e0 \u0111\u01b0\u1eddng cao v\u1eeba l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn c\u1ee7a $\\Delta MON$ <br\/> $\\Rightarrow HM = HN = OM = \\dfrac{MN}{2} = x$ (\u0111\u01b0\u1eddng trung tuy\u1ebfn b\u1eb1ng n\u1eeda c\u1ea1nh huy\u1ec1n) <br\/>X\u00e9t $\\Delta OMH$ c\u00f3: $\\widehat{H}=90^o$ <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pitago ta c\u00f3: <br\/> $OM^2=OH^2+MH^2\\\\\\Rightarrow 5=x^2+x^2\\\\ \\Rightarrow x =\\dfrac{\\sqrt{10}}{2}$ <br\/> Suy ra $MN = 2 HM = \\sqrt{10}$<\/span>"}]}],"id_ques":1146},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho \u0111\u01b0\u1eddng tr\u00f2n $(O; 5)$ v\u00e0 hai b\u00e1n k\u00ednh $OA$ v\u00e0 $OB$ vu\u00f4ng g\u00f3c v\u1edbi nhau. Tr\u00ean c\u00e1c b\u00e1n k\u00ednh $OA, OB$ l\u1ea7n l\u01b0\u1ee3t l\u1ea5y c\u00e1c \u0111i\u1ec3m $M$ v\u00e0 $N$ sao cho $OM = ON = \\sqrt{5}$. V\u1ebd d\u00e2y $CD$ \u0111i qua $M$ v\u00e0 $N$ ($M$ n\u1eb1m gi\u1eefa $C$ v\u00e0 $N$). <br\/><b> C\u00e2u 2: <\/b> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? <\/span>","select":["A. $CM = MN$ ","B. $MN = ND$","C. $CM = ND$","D. C\u1ea3 ba \u0111\u00e1p \u00e1n tr\u00ean \u0111\u1ec1u \u0111\u00fang"],"explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai7/lv3/img\/H922_K7.jpg' \/><\/center> <span class='basic_left'>Ta c\u00f3 $OH\\bot CD$ (theo c\u00e1ch v\u1ebd c\u00e2u 1) <br\/> $\\Rightarrow HC = HD$ (quan h\u1ec7 vu\u00f4ng g\u00f3c gi\u1eefa \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/> $MH = HN$ (theo c\u00e2u 1) $\\Rightarrow CM = ND$ <br\/>X\u00e9t $\\Delta OCH $ c\u00f3: $\\widehat{H}=90^o$ <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pitago ta c\u00f3: <br\/> $OC^2= OH^2 + CH^2 \\\\ \\Rightarrow 25=CH^2+ \\dfrac{5}{2} \\\\ \\Rightarrow CH =\\dfrac{3\\sqrt{10}}{2}$ <br\/> Suy ra $CM = CH - MH =$$\\dfrac {3\\sqrt{10}}{2} - \\dfrac{\\sqrt{10}}{2}$$=\\sqrt{10}$ <br\/> $\\Rightarrow CM = ND =\\sqrt{10}$ <br\/> M\u00e0 $MN =\\sqrt{10}$ (theo c\u00e2u 1) <br\/> $\\Rightarrow CM = MN = ND$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span>","column":2}]}],"id_ques":1147},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<span class='basic_left'> Trong c\u00e1c d\u00e2y \u0111i qua m\u1ed9t \u0111i\u1ec3m n\u1eb1m trong \u0111\u01b0\u1eddng tr\u00f2n, d\u00e2y ng\u1eafn nh\u1ea5t l\u00e0 d\u00e2y vu\u00f4ng g\u00f3c v\u1edbi \u0111\u01b0\u1eddng k\u00ednh \u0111i qua \u0111i\u1ec3m \u0111\u00f3 <\/span>","select":["A. \u0110\u00fang","B. Sai"],"explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai7/lv3/img\/H922_K1.jpg' \/><\/center> <span class='basic_left'> Gi\u1ea3 s\u1eed, \u0111i\u1ec3m $M$ n\u1eb1m trong \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m $O.$ D\u00e2y $AB$ \u0111i qua $M$ v\u00e0 vu\u00f4ng g\u00f3c $OM$ t\u1ea1i $M$ <br\/> G\u1ecdi $CD$ l\u00e0 d\u00e2y b\u1ea5t k\u00ec (kh\u00e1c d\u00e2y $AB$) \u0111i qua \u0111i\u1ec3m $M,$ k\u1ebb $OI \\bot CD$ <br\/>$\\Delta OMI$ c\u00f3: $\\widehat{I}=90^o$ $\\Rightarrow OM > OI$ (trong tam gi\u00e1c vu\u00f4ng c\u1ea1nh huy\u1ec1n l\u1edbn h\u01a1n c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow AB < CD$ (\u0111\u1ecbnh l\u00ed v\u1ec1 kho\u1ea3ng c\u00e1ch t\u1eeb t\u00e2m \u0111\u1ebfn d\u00e2y) <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0111\u00fang <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span>","column":2}]}],"id_ques":1148},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho \u0111\u01b0\u1eddng tr\u00f2n $O$ b\u00e1n k\u00ednh $R,$ hai d\u00e2y cung $AB$ v\u00e0 $CD.$ C\u00e1c tia $BA$ v\u00e0 $DC$ c\u1eaft nhau t\u1ea1i \u0111i\u1ec3m $M$ n\u1eb1m b\u00ean ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n. H\u1ea1 $OH \\bot AB$ t\u1ea1i $H,$ $OK \\bot CD$ t\u1ea1i $K.$ <br\/> <b> C\u00e2u 1: <\/b> N\u1ebfu $AB = CD.$ So s\u00e1nh $MA$ v\u00e0 $MC$ <\/span>","select":["A. $MA > MC$","B. $MA < MC$","C. $MA = MC$"],"explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn <\/span><br\/> Ta \u0111i x\u00e9t m\u1ed1i quan h\u1ec7 gi\u1eefa hai tam gi\u00e1c $MOH$ v\u00e0 $MOK$ \u0111\u1ec3 so s\u00e1nh $MA$ v\u00e0 $MC.$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai7/lv3/img\/H922_K3.jpg' \/><\/center> <br\/> <span class='basic_left'> Ta c\u00f3: $OH \\bot AB,$ $OK \\bot CD$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow HA=HB = \\dfrac{1}{2}AB\\\\ CK = KD= \\dfrac{1}{2} CD$ (quan h\u1ec7 vu\u00f4ng g\u00f3c gi\u1eefa \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/>N\u1ebfu $ AB = CD$ $\\Rightarrow HA = CK$ <br\/> V\u00e0 $OH = OK$ (\u0111\u1ecbnh l\u00ed v\u1ec1 kho\u1ea3ng c\u00e1ch t\u1eeb t\u00e2m \u0111\u1ebfn d\u00e2y) <br\/> X\u00e9t $\\Delta MOH$ v\u00e0 $\\Delta MOK$ c\u00f3: <br\/> $\\begin{align}& \\widehat{H}=\\widehat{K} = 90^o\\\\ & OH = OK \\\\& OM \\,\\, \\text{chung} \\\\ \\end{align}$ <br\/> $\\Rightarrow \\Delta MOH =\\Delta MOK $ (c\u1ea1nh huy\u1ec1n - c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> Suy ra $MH = MK$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow MH - AH = MK - CK$ <br\/> $\\Rightarrow MA = MC$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span>","column":3}]}],"id_ques":1149},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'>Cho \u0111\u01b0\u1eddng tr\u00f2n $O$ b\u00e1n k\u00ednh $R,$ hai d\u00e2y cung $AB$ v\u00e0 $CD.$ C\u00e1c tia $BA$ v\u00e0 $DC$ c\u1eaft nhau t\u1ea1i \u0111i\u1ec3m $M$ n\u1eb1m b\u00ean ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n. H\u1ea1 $OH \\bot AB$ t\u1ea1i $H,$ $OK \\bot CD$ t\u1ea1i $K.$ <br\/><b> C\u00e2u 2: <\/b> Bi\u1ebft $AB > CD.$ Ch\u1ee9ng minh $HM > KM$ <\/span>","title_trans":"S\u1eafp x\u1ebfp th\u1ee9 t\u1ef1 c\u00e1c b\u01b0\u1edbc ch\u1ee9ng minh b\u00e0i to\u00e1n ","temp":"sequence","correct":[[[4],[3],[2],[1],[5]]],"list":[{"point":10,"image":"https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai7/lv3/img\/H922_K4.jpg","left":["V\u00ec $AB > CD$ n\u00ean $OH < OK$ $\\Rightarrow OH^2 < OK^2$ $\\Rightarrow HM^2 > KM^2$","Suy ra $OH^2+HM^2= OK^2 +KM^2$","$OH^2+HM^2=OM^2$; $OK^2 +KM^2=OM^2$ ","\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pitago l\u1ea7n l\u01b0\u1ee3t v\u00e0o c\u00e1c tam gi\u00e1c vu\u00f4ng $MOH$ v\u00e0 $MOK$ ta c\u00f3: ","Suy ra $HM > KM$"],"top":80,"explain":"<span class='basic_left'> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pitago l\u1ea7n l\u01b0\u1ee3t v\u00e0o c\u00e1c tam gi\u00e1c vu\u00f4ng $MOH$ v\u00e0 $MOK$ ta c\u00f3: <br\/> $OH^2+HM^2=OM^2$ <br\/> $OK^2 +KM^2=OM^2$ <br\/> Suy ra $OH^2+HM^2= OK^2 +KM^2$ <br\/> V\u00ec $AB > CD$ n\u00ean $OH < OK$ $\\Rightarrow OH^2 < OK^2$ $\\Rightarrow HM^2 > KM^2$<br\/> Suy ra $HM > KM$ <\/span>"}]}],"id_ques":1150}],"lesson":{"save":0,"level":3}}