{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{25}{3}$","B. $\\dfrac{20}{3}$","C. $\\dfrac{22}{3}$"],"ques":"<span class='basic_left'>Cho \u0111\u01b0\u1eddng tr\u00f2n $(O; 5\\, cm)$ v\u00e0 \u0111i\u1ec3m $A$ n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n. K\u1ebb hai ti\u1ebfp tuy\u1ebfn $AM, AN$ \u0111\u1ebfn \u0111\u01b0\u1eddng tr\u00f2n ($M, N$ l\u00e0 c\u00e1c ti\u1ebfp \u0111i\u1ec3m). Bi\u1ebft $MN = 8 cm.$ \u0110o\u1ea1n th\u1eb3ng $OA$ d\u00e0i bao nhi\u00eau $cm?$ <br\/><br\/><\/span>","hint":"S\u1eed d\u1ee5ng h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv3/img\/h924_k1.1.png' \/><\/center><span class='basic_left'>G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $MN$ v\u00e0 $OA$ <br\/> $OM = ON = R; AM = AN$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> $\\Rightarrow OA$ l\u00e0 trung tr\u1ef1c c\u1ee7a $MN$ <br\/> $\\Rightarrow MI\\bot OA;$$MI=\\dfrac{MN}{2}$$=\\dfrac{8}{2}=4\\left( cm \\right)$ <br\/>X\u00e9t $\\Delta MIO$ vu\u00f4ng t\u1ea1i $I$ <br\/> $OI^2 + MI^2=OM^2$ (\u0111\u1ecbnh l\u00ed Pitago) <br\/> $\\Rightarrow OI=\\sqrt{OM^2 - MI^2} <br\/> $ <br\/> $=\\sqrt{25-16}<br\/>=3\\left( cm \\right)$ <br\/> X\u00e9t $\\Delta OMA$ vu\u00f4ng t\u1ea1i $M;$ $MI\\bot AO$ <br\/> $\\Rightarrow O{{M}^{2}}=OI.OA$ (h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Rightarrow OA=\\dfrac{OM^2}{OI}$$=\\dfrac{25}{3}\\left( cm \\right)$<\/span>"}]}],"id_ques":1201},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$, c\u00f3 $AB = 4cm, BC = 5cm.$ B\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp tam gi\u00e1c $ABC$ b\u1eb1ng: <\/span>","select":["A. $2$","B. $1$","C. $0,5$","D. \u0110\u00e1p \u00e1n kh\u00e1c"],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv3/img\/h924_k3.png' \/><\/center><span class='basic_left'> G\u1ecdi $(I;r)$ l\u00e0 \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp tam gi\u00e1c $ABC$ <br\/> $D, E, F$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 ti\u1ebfp \u0111i\u1ec3m c\u1ee7a $AC, AB, BC$ v\u1edbi $(I)$ <br\/> Ta c\u00f3: $AB, BC, AC$ l\u00e0 c\u00e1c ti\u1ebfp tuy\u1ebfn c\u1ee7a $(I)$ <br\/> $\\Rightarrow AE = AD;$$ BE = BF;$$ CF = CD$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> X\u00e9t $\\Delta ABC$ vu\u00f4ng t\u1ea1i $A$ <br\/> $\\Rightarrow AC=\\sqrt{B{{C}^{2}}-A{{B}^{2}}}$ (\u0111\u1ecbnh l\u00ed Pitago) <br\/> $=\\sqrt{25-14}=3\\left( cm \\right)$ <br\/> T\u1ee9 gi\u00e1c $AEID$ c\u00f3: <br\/> $\\widehat{A}=\\widehat{I}=\\widehat{D}={{90}^{o}}$ <br\/>$\\Rightarrow AEID$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft)<br\/> M\u00e0 $IE=ID=r$ <br\/> $\\Rightarrow $ AEID l\u00e0 h\u00ecnh vu\u00f4ng <br\/> $\\Rightarrow AE=IE=r$ <br\/> D\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c: $AB+AC-BC=2AE$ <br\/> $\\Rightarrow r=AE $$=\\dfrac{AB+AC-BC}{2} $ <br\/> $=\\dfrac{3+4-5}{2}$$=1 \\left( cm \\right)$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span>","column":4}]}],"id_ques":1202},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ c\u00f3 $BC = a, AB = c, AB = b$ ngo\u1ea1i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(I; r)$. Di\u1ec7n t\u00edch tam gi\u00e1c $ABC$ b\u1eb1ng: <\/span>","select":["A. $r(a+b+c)$","B. $2r(a+b+c)$","C. $\\dfrac{1}{2}r(a+b+c)$","D. $\\dfrac{abc}{r}$"],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv3/img\/h924_k5.png' \/><\/center><span class='basic_left'> Ta c\u00f3: ${{S}_{AIB}}=\\dfrac{1}{2}DI.AB$ $=\\dfrac{1}{2}r.c $ <br\/> ${{S}_{BIC}}=\\dfrac{1}{2}IF.BC$ $=\\dfrac{1}{2}r.a $ <br\/> $ {{S }_{AIC}}=\\dfrac{1}{2}IE.AC$ $=\\dfrac{1}{2}r.b $ <br\/> $ {{S}_{ABC}}={{S}_{AIB}}$$+{{S}_{BIC}}+{{ S }_{AIC}}$ <br\/> $=\\dfrac{1}{2}r\\left( a+b+c \\right) $ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span>","column":2}]}],"id_ques":1203},{"time":24,"part":[{"title":"<span class='basic_left'>Cho tam gi\u00e1c $MNP$, \u0111\u01b0\u1eddng tr\u00f2n $(J)$ b\u00e0ng ti\u1ebfp trong g\u00f3c $M,$ ti\u1ebfp x\u00fac v\u1edbi c\u00e1c tia $MN$ v\u00e0 $MP$ theo th\u1ee9 t\u1ef1 t\u1ea1i $E$ v\u00e0 $F$. Cho $NP = m, MP =n, MN = p$. <br\/> <b> C\u00e2u 1: <\/b> Khi \u0111\u00f3:<\/span>","title_trans":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","temp":"true_false","correct":[["t","f","f","t"]],"list":[{"point":10,"image":"https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv3/img\/h924_k6.png","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["$ME =MF=\\dfrac{m+n+p}{2}$ ","$NE =\\dfrac{-m+n+p}{2}$","$ME = PF=\\dfrac{m+n-p}{2}$","$PF=\\dfrac{m-n+p}{2}$"],"explain":["<span class='basic_left'>\u0110\u00fang, ta c\u00f3 $ME=MF;NE=ND;PD=PF$ (t\u00ednh ch\u1ea5t)<br\/>$ME+MF$$=MN+NE$$+MP+PF$ <br\/> $=MN+MP$$+ND+DP$ <br\/> $=m+n+p$ <br\/> $\\Rightarrow ME=MF$$=\\dfrac{m+n+p}{2}$ <\/span>","<span class='basic_left'>Sai, v\u00ec $NE=ME-MN$ <br\/> $=\\dfrac{m+n+p}{2}-p$ <br\/> $=\\dfrac{m+n-p}{2}$<\/span>","<span class='basic_left'> Sai, v\u00ec $PF=MF-MP$ <br\/> $=\\dfrac{m+n+p}{2}-n$ <br\/> $=\\dfrac{m+p-n}{2}$<\/span> "]}]}],"id_ques":1204},{"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":[[["12"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<span class='basic_left'> Cho tam gi\u00e1c $MNP$, \u0111\u01b0\u1eddng tr\u00f2n $(J)$ b\u00e0ng ti\u1ebfp trong g\u00f3c $M$, ti\u1ebfp x\u00fac v\u1edbi c\u00e1c tia $MN$ v\u00e0 $MP$ theo th\u1ee9 t\u1ef1 t\u1ea1i $E$ v\u00e0 $F$. Cho $NP = 10cm, MP =6cm,$ $MN = 8cm$. <br\/> <b> C\u00e2u 2: <\/b> T\u00ednh b\u00e1n k\u00ednh $r$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n $(J)$. <br\/> <b> \u0110\u00e1p \u00e1n:<\/b> $r=$ _input_$(cm)$<\/span>","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv3/img\/h924_k9.1.png' \/><\/center><span class='basic_left'> Ta c\u00f3: $MN^2+MP^2=NP^2$ (do $6^2 + 8^2 = 10^2$)<br\/> $\\Rightarrow \\Delta MNP$ vu\u00f4ng t\u1ea1i $M$ (\u0111\u1ecbnh l\u00ed py-ta-go \u0111\u1ea3o) <br\/> D\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c t\u1ee9 gi\u00e1c $MEJF$ l\u00e0 h\u00ecnh vu\u00f4ng <br\/> $\\Rightarrow ME=JE=r$ <br\/> Theo c\u00e2u 1: $ME=\\dfrac{MN+MP+PN}{2}$ <br\/> $=\\dfrac{6+8+10}{2}$$=12\\, (cm)$ <br\/> <span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $12.$<\/span><\/span>"}]}],"id_ques":1205},{"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 n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n $(O)$, \u0111\u01b0\u1eddng k\u00ednh $AB$, \u0111i\u1ec3m $C$ di chuy\u1ec3n tr\u00ean n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n. K\u1ebb ti\u1ebfp tuy\u1ebfn $Ax$ v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n. \u0110\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $CAx$ giao v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $E$ v\u00e0 giao v\u1edbi $BC$ t\u1ea1i $D$. G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AC$ v\u00e0 $BE$.<br\/><br\/><b> C\u00e2u 1: <\/b> Tam gi\u00e1c $ABD$ l\u00e0 tam gi\u00e1c g\u00ec?<\/span>","select":["A. Vu\u00f4ng","B. C\u00e2n","C. \u0110\u1ec1u"],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv3/img\/h924_k12.png' \/><\/center><span class='basic_left'>Ta c\u00f3: $\\widehat{A_1} = \\widehat{A_2}$ (gi\u1ea3 thi\u1ebft) (1) <br\/> $OA = OE = OB$ (c\u00f9ng b\u1eb1ng b\u00e1n k\u00ednh $(O)$) <br\/> $\\Rightarrow \\Delta AEB$ vu\u00f4ng t\u1ea1i $E$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Rightarrow\\widehat{A_1} = \\widehat{B_1}$ (c\u00f9ng ph\u1ee5 $\\widehat{EAB}$) (2) <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 $\\widehat{xAC} = \\widehat{ABC}$ (c\u00f9ng ph\u1ee5 $\\widehat{CAB}$) <br\/> $\\Rightarrow \\widehat{A_1} + \\widehat{2_1} = \\widehat{B_1} + \\widehat{B_2}$ <br\/> $\\Rightarrow \\widehat{A_2} = \\widehat{B_2}$ (3) <br\/> T\u1eeb (1), (2), (3) $\\Rightarrow \\widehat{B_1} = \\widehat{B_2} $ <br\/> $\\Rightarrow BE$ l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c $CBA$ <br\/> $\\Rightarrow BE$ v\u1eeba l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c v\u1eeba l\u00e0 \u0111\u01b0\u1eddng cao n\u00ean tam gi\u00e1c $DBA$ c\u00e2n t\u1ea1i $B.$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span>","column":3}]}],"id_ques":1206},{"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 n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n $(O)$, \u0111\u01b0\u1eddng k\u00ednh $AB$, \u0111i\u1ec3m $C$ di chuy\u1ec3n tr\u00ean n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n. K\u1ebb ti\u1ebfp tuy\u1ebfn $Ax$ v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n. \u0110\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $CAx$ giao v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $E$ v\u00e0 giao v\u1edbi $BC$ t\u1ea1i $D$. G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AC$ v\u00e0 $BE$.<br\/><br\/><b> C\u00e2u 2: <\/b> Khi $C$ di chuy\u1ec3n tr\u00ean $(O)$ th\u00ec qu\u1ef9 t\u00edch \u0111i\u1ec3m $D$ l\u00e0:<\/span>","select":["A. M\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng","B. M\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n","C. M\u1ed9t cung tr\u00f2n"],"hint":"C\u00e1ch x\u00e1c \u0111\u1ecbnh qu\u1ef9 t\u00edch: L\u1ea5y \u0111i\u1ec3m $C$ t\u1ea1i hai v\u1ecb tr\u00ed kh\u00e1c nhau, x\u00e1c \u0111\u1ecbnh c\u00e1c \u0111i\u1ec3m $D'$ v\u00e0 $D''.$ Th\u00f4ng qua ba \u0111i\u1ec3m $D, D', D''$ x\u00e1c \u0111\u1ecbnh qu\u1ef9 t\u00edch \u0111i\u1ec3m $D$","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv3/img\/h924_k12.1.png' \/><\/center><span class='basic_left'>Theo c\u00e2u 1: $\\Delta ABD$ c\u00e2n t\u1ea1i $B$ <br\/> $\\Rightarrow BD=BA$ <br\/> M\u00e0 $BA$ kh\u00f4ng \u0111\u1ed5i n\u00ean $BD$ kh\u00f4ng \u0111\u1ed5i <br\/> V\u1eady qu\u1ef9 t\u00edch \u0111i\u1ec3m $D$ l\u00e0 \u0111\u01b0\u1eddng tr\u00f2n $(B; BA)$ <br\/> <b>Gi\u1edbi h\u1ea1n: <\/b> Khi $C$ tr\u00f9ng v\u1edbi $A$ th\u00ec $D$ tr\u00f9ng v\u1edbi $A$ <br\/> Khi $C$ tr\u00f9ng v\u1edbi $B$ th\u00ec $D$ tr\u00f9ng v\u1edbi $D'$. V\u1eady $D$ ch\u1ea1y tr\u00ean $\\overset\\frown{AD\u2019}$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n $(B; BA)$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span> <br\/> <b> Ch\u00fa \u00fd: <\/b> Khi l\u00e0m b\u00e0i to\u00e1n qu\u1ef9 t\u00edch ta c\u1ea7n ph\u1ea3i ch\u00fa \u00fd gi\u1edbi h\u1ea1n c\u1ee7a qu\u1ef9 t\u00edch. Khi $C$ b\u1ecb gi\u1edbi h\u1ea1n th\u00ec qu\u1ef9 t\u00edch $D$ c\u0169ng b\u1ecb gi\u1edbi h\u1ea1n <\/span>","column":3}]}],"id_ques":1207},{"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":[[["180"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<span class='basic_left'><br\/>Cho n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n $(O)$, \u0111\u01b0\u1eddng k\u00ednh $AB$, \u0111i\u1ec3m $C$ di chuy\u1ec3n tr\u00ean n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n. K\u1ebb ti\u1ebfp tuy\u1ebfn $Ax$ v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n. \u0110\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $CAx$ giao v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $E$ v\u00e0 giao v\u1edbi $BC$ t\u1ea1i $D$. G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AC$ v\u00e0 $BE$.<br\/><br\/><b> C\u00e2u 3: <\/b> Khi \u0111\u00f3: $\\widehat{xAI}+\\widehat{AID}$ l\u00e0 _input_ $^o$<\/span>","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv3/img\/h924_k12.png' \/><\/center><span class='basic_left'>Ta c\u00f3: $\\widehat{AEB}=\\widehat{ACB}={{90}^{o}}$ (theo c\u00e2u 1) <br\/> $\\Rightarrow BE\\bot AD;AC\\bot BD$ <br\/> $\\Rightarrow I$ l\u00e0 tr\u1ef1c t\u00e2m tam gi\u00e1c $\\Rightarrow DI\\bot AB;$ <br\/>M\u00e0 $Ax\\bot AB$ (t\u00ednh ch\u1ea5t) <br\/> $\\Rightarrow DI\/\/Ax$ <br\/> $\\Rightarrow \\widehat{xAI}+\\widehat{AID}={{180}^{o}}$ (t\u1ed5ng hai g\u00f3c trong c\u00f9ng ph\u00eda) <br\/> <span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $180.$<\/span><\/span>"}]}],"id_ques":1208},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["DN","ND"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<span class='basic_left'><br\/>Cho n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n $(O)$, \u0111\u01b0\u1eddng k\u00ednh $AB$. \u0110i\u1ec3m $M$ di chuy\u1ec3n tr\u00ean n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n. Ti\u1ebfp tuy\u1ebfn t\u1ea1i $M$ v\u00e0 $B$ c\u1ee7a n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n c\u1eaft nhau t\u1ea1i $D$. Qua $O$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi $MB$, c\u1eaft ti\u1ebfp tuy\u1ebfn t\u1ea1i $M$ \u1edf $C$ v\u00e0 c\u1eaft ti\u1ebfp tuy\u1ebfn t\u1ea1i $B$ \u1edf $N$.<br\/><b>C\u00e2u 1:<\/b> Khi \u0111\u00f3 ta c\u00f3 $DC = $ _input_ <\/span>","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv3/img\/h924_k13.png' \/><\/center><span class='basic_left'>Ta c\u00f3: $DM = DB$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> $\\Rightarrow \\Delta DMB$ c\u00e2n (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow \\widehat{{{M}_{1}}} = \\widehat{{{B}_{1}}}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> $MB \/\/ CN$ (gi\u1ea3 thi\u1ebft) $\\Rightarrow \\widehat{{{M}_{1}}} = \\widehat{{{C}_{1}}}$; $\\widehat{{{B}_{1}}} = \\widehat{N}$ (hai g\u00f3c \u0111\u1ed3ng v\u1ecb) <br\/> $\\Rightarrow \\widehat{{{C}_{1}}} = \\widehat{N}$ <br\/> $\\Rightarrow \\Delta CND$ c\u00e2n t\u1ea1i $D$ <br\/> $\\Rightarrow DC=DN$ <br\/> <span class='basic_pink'>V\u1eady \u0111o\u1ea1n th\u1eb3ng c\u1ea7n \u0111i\u1ec1n l\u00e0 $DN.$ <\/span><\/span>"}]}],"id_ques":1209},{"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":[[["25"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<span class='basic_left'><br\/>Cho n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n $(O)$, \u0111\u01b0\u1eddng k\u00ednh $AB$. \u0110i\u1ec3m $M$ di chuy\u1ec3n tr\u00ean n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n. Ti\u1ebfp tuy\u1ebfn t\u1ea1i $M$ v\u00e0 $B$ c\u1ee7a n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n c\u1eaft nhau t\u1ea1i $D$. Qua $O$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi $MB$, c\u1eaft ti\u1ebfp tuy\u1ebfn t\u1ea1i $M$ \u1edf $C$ v\u00e0 c\u1eaft ti\u1ebfp tuy\u1ebfn t\u1ea1i $B$ \u1edf $N$.<br\/><b>C\u00e2u 2:<\/b> Bi\u1ebft $\\widehat{CAB}=90^o; R = 5\\,(cm)$, khi \u0111\u00f3 $AC.BD$ = _input_ <\/span>","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv3/img\/h924_k13.1.png' \/><\/center><span class='basic_left'> Do $\\widehat{CAB}=90^o$ (gi\u1ea3 thi\u1ebft); $ A\\in (O)$ <br\/> $\\Rightarrow AC$ l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a $(O)$ <br\/> $\\Rightarrow \\widehat{{{O}_{1}}}=\\widehat{{{O}_{3}}}$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/>Ta c\u00f3 $OD$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{MOB}$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> $\\Rightarrow \\widehat{{{O}_{4}}}=\\widehat{{{O}_{5}}}$ (t\u00ednh ch\u1ea5t ph\u00e2n gi\u00e1c) <br\/> $\\Rightarrow \\widehat{COD}=\\widehat{{{O}_{3}}}+\\widehat{{{O}_{4}}}$$=\\dfrac{1}{2}\\widehat{AOB}$$={{90}^{o}}$ <br\/>X\u00e9t $\\Delta COD$ vu\u00f4ng t\u1ea1i $O;$ $OM\\bot CD$ <br\/> $\\Rightarrow O{{M}^{2}}=CM.MD$ (h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng)<br\/>$\\Rightarrow 25=AC.BD$ <br\/> <br\/> <span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $25.$<\/span><\/span>"}]}],"id_ques":1210}],"lesson":{"save":0,"level":3}}