{"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'>Cho tam gi\u00e1c $\\Delta ABC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m $O$. Bi\u1ebft $\\widehat{A} = 40^o;\\, \\widehat{B} = 60^o$. K\u1ebb $OH,\\,OK,\\,OI$ l\u1ea7n l\u01b0\u1ee3t vu\u00f4ng g\u00f3c v\u1edbi c\u00e1c c\u1ea1nh $AB,\\, AC,\\, BC$. So s\u00e1nh c\u00e1c \u0111o\u1ea1n $OH,\\, OK,\\,OI$ ta \u0111\u01b0\u1ee3c:<\/span>","select":["A. $OH>OK>OI$","B. $OK>OH>OI$","C. $OH>OI>OK$","D. $OI>OK>OH$"],"explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv3/img\/h926_K1.png' \/><\/center><br\/> X\u00e9t $\\Delta ABC$ c\u00f3: <br\/> $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^o$ (\u0111\u1ecbnh l\u00ed t\u1ed5ng ba g\u00f3c c\u1ee7a tam gi\u00e1c) <br\/> $\\Rightarrow 40^o + 60^o + \\widehat{C} = 180^o$ <br\/> $\\Rightarrow \\widehat{C}= 180^o - 40^o - 60^o = {{80}^{o}}$ <br\/> $\\Rightarrow \\widehat{A}<\\widehat{B}<\\widehat{C}$ <br\/> $\\Rightarrow BC < AC < AB$ (\u0111\u1ecbnh l\u00ed v\u1ec1 g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n) <br\/> $\\Rightarrow OI > OK > OH$ (\u0111\u1ecbnh l\u00ed kho\u1ea3ng c\u00e1ch t\u1eeb t\u00e2m \u0111\u1ebfn d\u00e2y) <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D. <\/span><\/span>","column":2}]}],"id_ques":1261},{"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 ba \u0111\u01b0\u1eddng tr\u00f2n $(O;R)$, $(O';R')$, $(I;r)$. Trong \u0111\u00f3 $(O)$ v\u00e0 $(O')$ ti\u1ebfp x\u00fac ngo\u00e0i t\u1ea1i $A$. $(O)$ v\u00e0 $(I)$ ti\u1ebfp x\u00fac trong t\u1ea1i $A$ ($R > R' > r$). Trong c\u00e1c k\u1ebft lu\u1eadn sau, k\u1ebft lu\u1eadn n\u00e0o kh\u00f4ng \u0111\u00fang?<\/span>","select":["A. $A, O, O'$ th\u1eb3ng h\u00e0ng","B. $A, O', I$ th\u1eb3ng h\u00e0ng","C. $A, O, O', I$ kh\u00f4ng th\u1eb3ng h\u00e0ng","D. $OO' = R + R'$ v\u00e0 $OI = R - r$"],"explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv3/img\/h926_K2.png' \/><\/center> Ta c\u00f3: $(O)$ v\u00e0 $(O')$ ti\u1ebfp x\u00fac ngo\u00e0i t\u1ea1i $A$ n\u00ean $O, O', A$ th\u1eb3ng h\u00e0ng v\u00e0 $OO'= R + R'$<br\/> $(O)$ v\u00e0 $(I)$ ti\u1ebfp x\u00fac trong t\u1ea1i $A$ n\u00ean $O, I, A$ th\u1eb3ng h\u00e0ng v\u00e0 $OI = R - r$ <br\/> Suy ra $A, O, O', I$ th\u1eb3ng h\u00e0ng <br\/> Suy ra $A, O', I$ th\u1eb3ng h\u00e0ng <br\/> V\u1eady c\u00e1c \u0111\u00e1p \u00e1n A, B, D \u0111\u00fang, \u0111\u00e1p \u00e1n C sai<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C. <\/span><\/span>","column":2}]}],"id_ques":1262},{"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 hai \u0111\u01b0\u1eddng tr\u00f2n $(O; 10cm)$ v\u00e0 $(I; 17cm)$ c\u1eaft nhau t\u1ea1i $M$ v\u00e0 $N$. Bi\u1ebft kho\u1ea3ng c\u00e1ch g\u1eefa hai t\u00e2m l\u00e0 $21 cm$. T\u00ednh \u0111\u1ed9 d\u00e0i d\u00e2y cung $MN$. <\/span>","select":["A. $ 15\\, cm$","B. $ 8\\, cm$","C. $ 7\\, cm$","D. $ 16\\, cm$"],"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv3/img\/h926_K4.1.png' \/><\/center> G\u1ecdi $H$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $MN$ v\u00e0 $OI$ <br\/> $\\Rightarrow OI\\bot MN$; $MH=HN$ (\u0111\u1ecbnh l\u00ed \u0111\u01b0\u1eddng n\u1ed1i t\u00e2m) <br\/> \u0110\u1eb7t $OH=x$ $\\Rightarrow HI=21-x$ <br\/>X\u00e9t $\\Delta MHO$ vu\u00f4ng t\u1ea1i $H$ <br\/> $OM^2 = OH^2 + MH^2$ (\u0111\u1ecbnh l\u00ed Pitago) <br\/>$\\Rightarrow MH^2 = OM^2 - OH^2$ <br\/>X\u00e9t $\\Delta MHI$ vu\u00f4ng t\u1ea1i $H$ <br\/> $IM^2 = MH^2 + IH^2$ (\u0111\u1ecbnh l\u00ed Pitago) <br\/> $\\Rightarrow MH^2 = IM^2 - HI^2$ <br\/> $\\Rightarrow OM^2 - OH^2=IM^2 - HI^2$ <br\/> $\\Leftrightarrow {10^2 - x^2}={{17}^{2}}-{{\\left( 21-x \\right)}^{2}}$ <br\/> $\\Leftrightarrow 100 - x^2=289-441+42x-{{x}^{2}}$ <br\/> $\\Leftrightarrow 42x=252$ <br\/> $\\Leftrightarrow x=6 (cm)$ <br\/> $\\Rightarrow MH=\\sqrt{{{10}^{2}}-{{6}^{2}}}= 8\\, (cm)$ <br\/> $MN = 2MH = 16\\, (cm)$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D. <\/span><\/span>","column":4}]}],"id_ques":1263},{"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 hai \u0111\u01b0\u1eddng tr\u00f2n $(O; 8cm)$ v\u00e0 $(I; 6cm)$ ti\u1ebfp x\u00fac ngo\u00e0i t\u1ea1i $A$. $MN$ l\u00e0 m\u1ed9t ti\u1ebfp tuy\u1ebfn chung ngo\u00e0i c\u1ee7a $(O)$ v\u00e0 $(I)$. \u0110\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng $MN$ l\u00e0:<\/span>","select":["A. $8\\, cm$","B. $9\\sqrt{3}\\, cm$","C. $9\\sqrt{2}\\, cm$","D. $8\\sqrt{3}\\, cm$"],"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv3/img\/h926_K5.png' \/><\/center> K\u1ebb ti\u1ebfp tuy\u1ebfn chung $AE$ c\u1ee7a hai \u0111\u01b0\u1eddng tr\u00f2n <br\/> Ta c\u00f3 $OE$ v\u00e0 $IE$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{MEA}$ v\u00e0 $\\widehat{NEA}$ <br\/> $\\Rightarrow \\left\\{ \\begin{align} & \\widehat{OEA}=\\dfrac{1}{2}\\widehat{MEA} \\\\ & \\widehat{IEA}=\\dfrac{1}{2}\\widehat{NEA} \\\\ \\end{align} \\right.$ (t\u00ednh ch\u1ea5t ph\u00e2n gi\u00e1c) <br\/> $\\Rightarrow \\widehat{OEI}= \\widehat{OEA}+\\widehat{IEA}=\\dfrac{1}{2}\\left( \\widehat{MEA}+\\widehat{NEA} \\right)=\\dfrac{1}{2}{{180}^{o}}={{90}^{o}}$ <br\/>X\u00e9t $\\Delta OEI$ vu\u00f4ng t\u1ea1i $E$ <br\/> $A{{E}^{2}}=AO.AI$ (h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng) <br\/> $=8.6=48$ <br\/> $\\Rightarrow AE=4\\sqrt{3}\\,(cm)$ <br\/>M\u00e0 $EM=EN=EA$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> $\\Rightarrow MN=2AE=8\\sqrt{3}\\,(cm)$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span>","column":4}]}],"id_ques":1264},{"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 tam gi\u00e1c \u0111\u1ec1u $MNP$ ngo\u1ea1i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n b\u00e1n k\u00ednh $4 cm$. <br\/> <b> C\u00e2u 1: <\/b> C\u1ea1nh c\u1ee7a tam gi\u00e1c \u0111\u1ec1u $MNP$ c\u00f3 \u0111\u1ed9 d\u00e0i l\u00e0: <\/span>","select":["A. $8\\, cm$","B. $ 6\\sqrt{3}\\, cm$","C. $ 4\\sqrt{3}\\, cm$","D. $ 8\\sqrt{3}\\, cm$"],"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv3/img\/h926_K6.png' \/><\/center> G\u1ecdi $H$ l\u00e0 trung \u0111i\u1ec3m $NP$, $O$ l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp tam gi\u00e1c $ MNP$ <br\/> $\\Delta MNP$ \u0111\u1ec1u $\\Rightarrow MH \\bot NP$ <br\/> T\u00e2m $ O$ tr\u00f9ng v\u1edbi tr\u1ecdng t\u00e2m tam gi\u00e1c <br\/> $\\Rightarrow MH=3OH$$=12\\,(cm)$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) <br\/> X\u00e9t $\\Delta MHN$ vu\u00f4ng t\u1ea1i $H$ <br\/> $sin\\widehat{MNH}=\\dfrac{MH}{MN}$ <br\/> $\\Rightarrow MN=\\dfrac{MH}{\\sin 60^o}$ $=\\dfrac{12}{\\dfrac{\\sqrt{3}}{2}}$$=8\\sqrt{3}\\,(cm)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span>","column":4}]}],"id_ques":1265},{"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":[[2]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho tam gi\u00e1c \u0111\u1ec1u $MNP$ ngo\u1ea1i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n b\u00e1n k\u00ednh $4 cm$. <br\/> <b> C\u00e2u 2: <\/b> G\u1ecdi $K$ l\u00e0 ti\u1ebfp \u0111i\u1ec3m c\u1ee7a $MN$ v\u00e0 $(O)$. T\u00ednh di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c $NKOH$. <\/span>","select":["A. $48\\sqrt{3}\\, cm^2$","B. $ 16\\sqrt{3}\\, cm^2$","C. $ 32\\sqrt{3}\\, cm$","D. $ 8\\sqrt{3}\\, cm^2$"],"hint":"D\u1ef1a v\u00e0o t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch","explain":"<span class='basic_left'><span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $NOH.$ <br\/> B\u01b0\u1edbc 2: Ch\u1ee9ng minh $\\Delta NOH=\\Delta NOK.$ Suy ra ${{S}_{NOH}}={{S}_{NOK}}.$ <br\/> B\u01b0\u1edbc 3: T\u00ednh di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c $NKOH$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv3/img\/h926_K6.1.png' \/><\/center> Ta c\u00f3: $MN = NP = 8\\sqrt{3}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c \u0111\u1ec1u) <br\/> $NH = \\dfrac{NP}{2}=4\\sqrt{3}\\,(cm)$ <br\/> $\\Rightarrow {S}_{NOH} = \\dfrac{1}{2} OH. NP$$ = \\dfrac{4.4\\sqrt{3}}{2} = 8\\sqrt{3}\\, (cm^2)$ <br\/> X\u00e9t hai tam gi\u00e1c vu\u00f4ng $\\Delta NOH$ v\u00e0 $\\Delta NOK$ c\u00f3: <br\/> $NO$ chung <br\/> $OH = OK = 4\\,(cm)$ <br\/> $\\Rightarrow\\Delta NOH=\\Delta NOK$ (c\u1ea1nh huy\u1ec1n v\u00e0 c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow {{S}_{NOH}}={{S}_{NOK}}$ <br\/> $\\Rightarrow {{S}_{NKOH}} = S_{OHN} + S_{OKN}$ $=2S_{OHN}= 2.8\\sqrt{3} = 16\\sqrt{3}\\,(cm^2)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span> <\/span>","column":4}]}],"id_ques":1266},{"time":24,"part":[{"title":"","title_trans":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","temp":"true_false","correct":[["f","t","f","t"]],"list":[{"point":10,"image":"","ques":"","col_name":["C\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00fang hay sai?","\u0110\u00fang","Sai"],"arr_ques":["<span class='basic_left'>\u0110\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi b\u00e1n k\u00ednh l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00f3 <\/span> ","<span class='basic_left'>N\u1ebfu $CA, CB$ l\u00e0 c\u00e1c ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n $(O)$ v\u1edbi $A, B$ l\u00e0 c\u00e1c ti\u1ebfp \u0111i\u1ec3m th\u00ec $OC$ l\u00e0 trung tr\u1ef1c c\u1ee7a $AB$<\/span>","<span class='basic_left'>Trong m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n, t\u1ed3n t\u1ea1i d\u00e2y cung c\u00f3 \u0111\u1ed9 d\u00e0i l\u1edbn h\u01a1n \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a n\u00f3<\/span>","<span class='basic_left'>\u0110\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a d\u00e2y cung l\u00e0 \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n ch\u1ee9a d\u00e2y cung \u1ea5y<\/span>"],"explain":["<span class='basic_left'> Sai, v\u00ec n\u1ebfu m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua m\u1ed9t \u0111i\u1ec3m c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi b\u00e1n k\u00ednh \u0111i qua \u0111i\u1ec3m \u1ea5y th\u00ec \u0111\u01b0\u1eddng th\u1eb3ng \u1ea5y l\u00e0 m\u1ed9t ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n<\/span>","\u0110\u00fang, v\u00ec <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv3/img\/h926_D3.2.png' \/><\/center><span class='basic_left'> Ta c\u00f3 $CA =CB $ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) $\\Rightarrow C$ thu\u1ed9c trung tr\u1ef1c c\u1ee7a $AB$ <br\/> $OB=OA=R$ $\\Rightarrow O$ thu\u1ed9c trung tr\u1ef1c c\u1ee7a $AB$ <br\/> $\\Rightarrow OC$ l\u00e0 trung tr\u1ef1c c\u1ee7a $AB$<\/span>","<span class='basic_left'>Sai, v\u00ec trong m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n, \u0111\u01b0\u1eddng k\u00ednh l\u00e0 d\u00e2y cung c\u00f3 \u0111\u1ed9 d\u00e0i l\u1edbn nh\u1ea5t<\/span>","<span class='basic_left'>\u0110\u00fang, v\u00ec trong m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n, \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a m\u1ed9t d\u00e2y \u0111i qua trung \u0111i\u1ec3m v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi d\u00e2y \u1ea5y th\u00ec l\u00e0 \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n <\/span>"]}]}],"id_ques":1267},{"time":9,"part":[{"time":3,"title":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ nh\u1ecdn. \u0110\u01b0\u1eddng tr\u00f2n t\u00e2m $(O:R)$, \u0111\u01b0\u1eddng k\u00ednh $BC$ c\u1eaft c\u1ea1nh $AB$ t\u1ea1i $M$ v\u00e0 c\u1eaft c\u1ea1nh $AC$ t\u1ea1i $N$. G\u1ecdi $H$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BN$ v\u00e0 $CM$, $K$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AH$ v\u00e0 $BC$. <br\/> <b> C\u00e2u 1: <\/b> Ch\u1ee9ng minh $AH$ vu\u00f4ng g\u00f3c v\u1edbi $BC$. <\/span>","title_trans":"S\u1eafp x\u1ebfp th\u1ee9 t\u1ef1 c\u00e1c b\u01b0\u1edbc ch\u1ee9ng minh","temp":"sequence","correct":[[[3],[2],[5],[1],[4]]],"list":[{"point":10,"image":"https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv3/img\/h926_K8.png","left":["$\\Rightarrow BN\\bot AC;$$CM\\bot AB$","Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1: $\\Rightarrow\\widehat{BNC}={{90}^{o}}$","Suy ra $ AH\\bot BC$","Ta c\u00f3: $OM = OB = OC =R$$\\Rightarrow\\widehat{BMC}={{90}^{o}}$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn) "," M\u00e0 $BN\\cap CM=H$ $\\Rightarrow H$ l\u00e0 tr\u1ef1c t\u00e2m $\\Delta ABC$"],"top":55,"explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv3/img\/h926_K8.png' \/><\/center> Ta c\u00f3: $OM = OB = OC = R$$\\Rightarrow\\widehat{BMC}={{90}^{o}}$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn) <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1: $\\Rightarrow\\widehat{BNC}={{90}^{o}}$ <br\/> $\\Rightarrow BN\\bot AC;$$CM\\bot AB$ <br\/> M\u00e0 $BN\\cap CM=H$ <br\/> $\\Rightarrow H$ l\u00e0 tr\u1ef1c t\u00e2m $\\Delta ABC$ <br\/> Suy ra $ AH\\bot BC.$ <\/span>"}]}],"id_ques":1268},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank","correct":[[["90"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$ nh\u1ecdn. \u0110\u01b0\u1eddng tr\u00f2n t\u00e2m $(O; R)$, \u0111\u01b0\u1eddng k\u00ednh $BC$ c\u1eaft c\u1ea1nh $AB$ t\u1ea1i $M$ v\u00e0 c\u1eaft c\u1ea1nh $AC$ t\u1ea1i $N$. G\u1ecdi $H$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BN$ v\u00e0 $CM$, $K$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AH$ v\u00e0 $BC$. <br\/> <b> C\u00e2u 2: <\/b> G\u1ecdi $E$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AH$. Khi \u0111\u00f3 s\u1ed1 \u0111o g\u00f3c $\\widehat{EMO}$ l\u00e0 _input_ $^o$<\/span>","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv3/img\/h926_K8.1.png' \/><\/center><span class='basic_left'> Ta c\u00f3: $\\widehat{{{H}_{1}}}=\\widehat{{{H}_{2}}}$ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh) <br\/> $\\widehat{{{H}_{1}}}+\\widehat{A_1}$$=\\widehat{{{H}_{2}}}+\\widehat{C_1}$$={{90}^{o}} $ (1) <br\/> $\\Rightarrow \\widehat{A_1}=\\widehat{C_1} $ <br\/> $OM=OC$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow \\Delta OMC $ c\u00e2n t\u1ea1i $O$ (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow \\widehat{C_1}=\\widehat{OMC} $ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow \\widehat{OMC}=\\widehat{A_1}$ (2) <br\/> V\u00ec $E$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AH$ <br\/> $\\Rightarrow EM=EH$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) $\\Rightarrow \\Delta EHM $ c\u00e2n t\u1ea1i E <br\/> $\\Rightarrow \\widehat{EMH}=\\widehat{{{H}_{1}}}$ (3) <br\/> T\u1eeb (1), (2), (3) $\\Rightarrow \\widehat{OMC}+ \\widehat{EMH}= \\widehat{A_1} + \\widehat{H_1} = 90^o$ <br\/> $\\Rightarrow\\widehat{EMO}=90^o$ <br\/> <span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $90.$<\/span><\/span>"}]}],"id_ques":1269},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u (>;<;=) th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank","correct":[[["="]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$ nh\u1ecdn. \u0110\u01b0\u1eddng tr\u00f2n t\u00e2m $(O; R)$, \u0111\u01b0\u1eddng k\u00ednh $BC$ c\u1eaft c\u1ea1nh $AB$ t\u1ea1i $M$ v\u00e0 c\u1eaft c\u1ea1nh $AC$ t\u1ea1i $N$. G\u1ecdi $H$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BN$ v\u00e0 $CM$, $K$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AH$ v\u00e0 $BK$. <br\/> <b> C\u00e2u 3: <\/b> Cho bi\u1ebft $\\sin\\widehat{BAC}=\\dfrac{\\sqrt{2}}{2}$. H\u00e3y so s\u00e1nh $AH$ v\u00e0 $BC$. <br\/><br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $AH $ _input_ $BC$ <\/span>","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv3/img\/h926_K8.1.png' \/><\/center><span class='basic_left'> Ta c\u00f3: $\\sin \\widehat{BAC}=\\dfrac{\\sqrt{2}}{2}$$\\Rightarrow \\widehat{BAC}={{45}^{o}}$ <br\/> $\\Rightarrow \\Delta AMC$ vu\u00f4ng c\u00e2n t\u1ea1i $M$<br\/> $\\Rightarrow MA=MC$ <br\/> X\u00e9t $\\Delta MAH$ v\u00e0 $\\Delta MCB$ <br\/> $\\widehat{MAH}=\\widehat{BMC}={{90}^{o}}$ <br\/> $MA=MC$ <br\/> $\\widehat{{{A}_{1}}}=\\widehat{{{C}_{1}}}$ (theo c\u00e2u 2) <br\/> $\\Rightarrow \\Delta MAH = \\Delta MCB$ (g.c.g) <br\/> $\\Rightarrow AH=BC$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> <span class='basic_pink'>V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0 d\u1ea5u $=$ <\/span><\/span>"}]}],"id_ques":1270}],"lesson":{"save":0,"level":3}}