{"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 $ABC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O), AB < AC$. Tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $A$ c\u1eaft $BC$ t\u1ea1i $D$ v\u00e0 c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n $(O)$ t\u1ea1i $M$ (kh\u00e1c $A$). Trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau, kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sai?","select":["A. $OM\\bot BC$ ","B. $\\widehat{BAM}=\\widehat{BCM}$ ","C. $DA.DM=DB.DC$ ","D. $\\widehat{ABM}=\\widehat{ACM}$"],"explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai15/lv3/img\/h933_K1.png' \/><\/center><br\/> Ta c\u00f3: $\\widehat{BAM}=\\widehat{CAM}$ (t\u00ednh ch\u1ea5t ph\u00e2n gi\u00e1c) <br\/> $\\Rightarrow \\overset\\frown{BM}=\\overset\\frown{CM}$ (g\u00f3c n\u1ed9i ti\u1ebfp b\u1eb1ng nhau ch\u1eafn hai cung b\u1eb1ng nhau) <br\/> $\\Rightarrow BM=CM$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y) <br\/> $\\Rightarrow M$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $BC$ <br\/> L\u1ea1i c\u00f3: $OA=OB$ (c\u00f9ng b\u1eb1ng b\u00e1n k\u00ednh $(O)$) <br\/> $\\Rightarrow O$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $BC$ <br\/> $\\Rightarrow OM$ l\u00e0 trung tr\u1ef1c c\u1ee7a $BC$ $\\Rightarrow OM\\bot BC$ <br\/> Suy ra \u0111\u00e1p \u00e1n A \u0111\u00fang <br\/> Ta c\u00f3: $\\widehat{BAM}=\\widehat{BCM}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn cung $\\overset\\frown{BM}$) <br\/> Suy ra \u0111\u00e1p \u00e1n B \u0111\u00fang<br\/> D\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c: $\\Delta ABD\\backsim \\Delta CMD\\,\\left( g.g \\right)$ <br\/> $\\Rightarrow \\dfrac{AD}{CD}=\\dfrac{BD}{DM}$ (c\u1eb7p t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng) <br\/> $\\Rightarrow AD.DM=CD.BD$ <br\/> Suy ra \u0111\u00e1p \u00e1n C \u0111\u00fang <br\/> L\u1ea1i c\u00f3: $\\widehat{CBM}=\\widehat{BCM}$ (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn hai cung b\u1eb1ng nhau) <br\/> M\u00e0 $AB < AC\\Rightarrow \\widehat{ABC} > \\widehat{ACB}$ (quan h\u1ec7 gi\u1eefa c\u1ea1nh v\u00e0 g\u00f3c trong tam gi\u00e1c) <br\/> $\\Rightarrow \\widehat{ABC}+\\widehat{CBM}>\\widehat{ACB}+\\widehat{BCM}\\,hay\\,\\widehat{ABM}>\\widehat{ACM}$ <br\/> Suy ra \u0111\u00e1p \u00e1n D sai. <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D <\/span><\/span>","column":2}]}],"id_ques":1491},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u ($+,-,.,:$ ho\u1eb7c $=$) th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["="],["+"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" <span class='basic_left'> Cho tam gi\u00e1c $ABC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O), AB < AC$. Tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $A$ c\u1eaft $BC$ t\u1ea1i $D$ v\u00e0 c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n $(O)$ t\u1ea1i $M$ (kh\u00e1c $A$). Khi \u0111\u00f3: $\\widehat{BMC}\\,\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\,\\widehat{ABC}\\,\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\,\\widehat{ACB}$","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai15/lv3/img\/h933_K1.png' \/><\/center> <br\/> Ta c\u00f3: $\\widehat{AMC}=\\widehat{ABC}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn cung $\\overset\\frown{AC}$) <br\/> $\\widehat{AMB}=\\widehat{ACB}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn cung $\\overset\\frown{AB}$) <br\/> $\\Rightarrow \\widehat{AMC}+\\widehat{AMB}=\\widehat{ABC}+\\widehat{ACB}$ <br\/> $\\Rightarrow \\widehat{BMC}=\\widehat{ABC}+\\widehat{ACB}$ <br\/> <span class='basic_pink'> V\u1eady c\u00e1c d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $=$ v\u00e0 $+$ <\/span><\/span> "}]}],"id_ques":1492},{"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 \u0111\u01b0\u1eddng tr\u00f2n $(O)$, d\u00e2y $AB$ song song v\u1edbi $CD$. Ch\u1ee9ng minh r\u1eb1ng $ABCD$ l\u00e0 h\u00ecnh thang c\u00e2n. B\u1ea1n An ch\u1ee9ng minh nh\u01b0 sau l\u00e0 \u0110\u00fang hay Sai? <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai15/lv3/img\/h933_K3.png' \/><\/center> <br\/> Ta c\u00f3: $AB\/\/CD$ $\\Rightarrow ABCD$ l\u00e0 h\u00ecnh thang (1) <br\/> $\\overset\\frown{AD}=\\overset\\frown{BC}$ (hai cung b\u1ecb ch\u1eafn b\u1edfi hai d\u00e2y song song) <br\/> $\\Rightarrow AD=BC$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y) (2) <br\/> T\u1eeb (1) v\u00e0 (2) suy ra $ABCD$ l\u00e0 h\u00ecnh thang c\u00e2n","select":["A. \u0110\u00fang ","B. Sai "],"explain":"<span class='basic_left'> D\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft h\u00ecnh thang c\u00e2n: <br\/> + H\u00ecnh thang c\u00e2n l\u00e0 h\u00ecnh thang c\u00f3 hai g\u00f3c k\u1ec1 m\u1ed9t c\u1ea1nh \u0111\u00e1y b\u1eb1ng nhau <br\/> + H\u00ecnh thang c\u00f3 hai \u0111\u01b0\u1eddng ch\u00e9o b\u1eb1ng nhau <br\/> Suy ra b\u1ea1n An ch\u1ee9ng minh ch\u01b0a \u0111\u00fang <br\/> Tr\u00ecnh b\u00e0y l\u1ea1i ta \u0111\u01b0\u1ee3c: <br\/> Ta c\u00f3: $AB\/\/CD$ $\\Rightarrow ABCD$ l\u00e0 h\u00ecnh thang (1) <br\/> $\\overset\\frown{AD}=\\overset\\frown{BC}$ (hai cung b\u1ecb ch\u1eafn b\u1edfi hai d\u00e2y song song) <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{AD}=\\text{s\u0111}\\overset\\frown{BC}$ <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{AD}+\\text{s\u0111}\\overset\\frown{DC}=\\text{s\u0111}\\overset\\frown{BC}+\\text{s\u0111}\\overset\\frown{CD}$ <br\/> $\\Rightarrow \\overset\\frown{ADC}=\\overset\\frown{BCD}\\Rightarrow \\widehat{B}=\\widehat{A}$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) suy ra $ABCD$ l\u00e0 h\u00ecnh thang c\u00e2n <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B <\/span><\/span>","column":2}]}],"id_ques":1493},{"time":24,"part":[{"time":3,"title":"Cho hai \u0111\u01b0\u1eddng tr\u00f2n $(O)$ v\u00e0 $(O\u2019)$ c\u1eaft nhau t\u1ea1i $A$ v\u00e0 $B$. M\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng c\u1eaft hai \u0111\u01b0\u1eddng tr\u00f2n $(O)$ v\u00e0 $(O\u2019)$ l\u1ea7n l\u01b0\u1ee3t t\u1ea1i b\u1ed1n \u0111i\u1ec3m $C, E, D, F$. Ch\u1ee9ng minh r\u1eb1ng: $\\widehat{CAF}+\\widehat{DBE}={{180}^{o}}$","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[2],[4],[1],[3]]],"list":[{"point":10,"left":["Trong $(O\u2019)$, ta c\u00f3: $\\widehat{DBA}=\\widehat{DFA}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn cung $\\overset\\frown{AD}$) (2) ","$\\Rightarrow \\widehat{CAF}+\\widehat{DBE}=\\widehat{CAF}+\\widehat{DBA}+\\widehat{ABE}$ <br\/> $=\\widehat{CAF}+\\widehat{ACE}+\\widehat{AFD}={{180}^{o}}$ (t\u1ed5ng ba g\u00f3c c\u1ee7a tam gi\u00e1c $ACF$)"," Trong $(O)$, ta c\u00f3: $\\widehat{ACE}=\\widehat{ABE}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn cung $\\overset\\frown{AE}$) (1) "," T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\widehat{DBA}+\\widehat{ABE}=\\widehat{ACE}+\\widehat{AFD}$ "],"top":100,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai15/lv3/img\/h933_K4.png' \/><\/center> <br\/> Trong $(O)$, ta c\u00f3: $\\widehat{ACE}=\\widehat{ABE}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn cung $\\overset\\frown{AE}$) (1) <br\/> Trong $(O\u2019)$, ta c\u00f3: $\\widehat{DBA}=\\widehat{DFA}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn cung $\\overset\\frown{AD}$) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\widehat{DBA}+\\widehat{ABE}=\\widehat{ACE}+\\widehat{AFD}$ <br\/> $\\Rightarrow \\widehat{CAF}+\\widehat{DBE}=\\widehat{CAF}+\\widehat{DBA}+\\widehat{ABE}$ <br\/> $=\\widehat{CAF}+\\widehat{ACE}+\\widehat{AFD}={{180}^{o}}$ (t\u1ed5ng ba g\u00f3c c\u1ee7a tam gi\u00e1c $ACF$) <\/span>"}]}],"id_ques":1494},{"time":24,"part":[{"time":3,"title":"Cho \u0111\u01b0\u1eddng tr\u00f2n $(O;R)$, d\u00e2y cung $AB$. Qua \u0111i\u1ec3m $I$ thu\u1ed9c d\u00e2y $AB \\,(IA < IB)$, v\u1ebd d\u00e2y $CD$ vu\u00f4ng g\u00f3c v\u1edbi $AB$, k\u1ebb \u0111\u01b0\u1eddng k\u00ednh $CE$. Ch\u1ee9ng minh r\u1eb1ng $AD^2+BC^2$ c\u00f3 gi\u00e1 tr\u1ecb kh\u00f4ng \u0111\u1ed5i.","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[3],[5],[1],[6],[4],[2]]],"list":[{"point":10,"left":[" M\u00e0 $AB\\bot CD$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow AB\/\/DE$ (t\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song) ","X\u00e9t $\\Delta CBE$ vu\u00f4ng t\u1ea1oi $B$ c\u00f3: $B{{C}^{2}}+B{{E}^{2}}=C{{E}^{2}}$ (\u0111\u1ecbnh l\u00ed Pitago) "," Ta c\u00f3: $\\widehat{CDE}=\\widehat{CBE}={{90}^{o}}$ (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n)","$\\Rightarrow B{{C}^{2}}+A{{D}^{2}}={{\\left( 2R \\right)}^{2}}$ (do $AD=BE$) <br\/> $\\Rightarrow B{{C}^{2}}+A{{D}^{2}}=4{{R}^{2}}$ (kh\u00f4ng \u0111\u1ed5i)"," $\\Rightarrow \\overset\\frown{AD}=\\overset\\frown{BE}$ (hai cung b\u1ecb ch\u1eafn gi\u1eefa hai d\u00e2y song song) <br\/> $\\Rightarrow AD=BE$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y) "," $\\widehat{CDE}={{90}^{o}}\\Rightarrow CD\\bot DE$ "],"top":100,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai15/lv3/img\/h933_K5.png' \/><\/center> <br\/> Ta c\u00f3: $\\widehat{CDE}=\\widehat{CBE}={{90}^{o}}$ (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\widehat{CDE}={{90}^{o}}\\Rightarrow CD\\bot DE$ <br\/> M\u00e0 $AB\\bot CD$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow AB\/\/DE$ (t\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song) <br\/> $\\Rightarrow \\overset\\frown{AD}=\\overset\\frown{BE}$ (hai cung b\u1ecb ch\u1eafn gi\u1eefa hai d\u00e2y song song) <br\/> $\\Rightarrow AD=BE$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y) <br\/> X\u00e9t $\\Delta CBE$ vu\u00f4ng t\u1ea1i $B$ c\u00f3: <br\/> $B{{C}^{2}}+B{{E}^{2}}=C{{E}^{2}}$ (\u0111\u1ecbnh l\u00ed Pitago) <br\/> $\\Rightarrow B{{C}^{2}}+A{{D}^{2}}={{\\left( 2R \\right)}^{2}}$ (do $AD=BE$) <br\/> $\\Rightarrow B{{C}^{2}}+A{{D}^{2}}=4{{R}^{2}}$ (kh\u00f4ng \u0111\u1ed5i) <\/span>"}]}],"id_ques":1495},{"time":24,"part":[{"time":3,"title":"Cho tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$, \u0111\u01b0\u1eddng cao $AD$. V\u1ebd \u0111\u01b0\u1eddng tr\u00f2n $(O)$ \u0111i qua $A$ v\u00e0 ti\u1ebfp x\u00fac v\u1edbi $BC$ t\u1ea1i $B$. \u0110\u01b0\u1eddng tr\u00f2n $(O)$ c\u1eaft $AD$ t\u1ea1i \u0111i\u1ec3m $H$ (kh\u00e1c $A$) v\u00e0 c\u1eaft $CH$ t\u1ea1i \u0111i\u1ec3m $K$ (kh\u00e1c $H$). Ch\u1ee9ng minh $AK =AC$","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[3],[1],[4],[2],[5]]],"list":[{"point":10,"left":[" $\\Rightarrow \\Delta ABH=\\Delta ACH\\,\\left( c.g.c \\right)$ ","Ta c\u00f3: $\\widehat{AKH}=\\widehat{ABH}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn $\\overset\\frown{AH}$) <br\/> $AD$ l\u00e0 \u0111\u01b0\u1eddng cao n\u00ean c\u0169ng l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $\\widehat{BAC}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) $\\Rightarrow\\widehat{BAH}=\\widehat{CAH}$ "," $\\Rightarrow \\widehat{ABH}=\\widehat{ACH}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow \\widehat{AKH}=\\widehat{ACH} $"," M\u1eb7t kh\u00e1c: X\u00e9t $\\Delta ABH$ v\u00e0 $\\Delta ACH$ c\u00f3: $\\left\\{ \\begin{align} & AB=AC \\,(\\text{gi\u1ea3 thi\u1ebft}) \\\\ & \\widehat{BAH}=\\widehat{CAH}\\left( \\text{ch\u1ee9ng minh tr\u00ean} \\right) \\\\ & AH \\text{chung} \\\\ \\end{align} \\right.$ "," $\\Rightarrow \\Delta AKC$ c\u00e2n t\u1ea1i $A$ (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <br\/> $\\Rightarrow AK=AC$ (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c c\u00e2n) "],"top":100,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai15/lv3/img\/h933_K6.png' \/><\/center> <br\/> Ta c\u00f3: $\\widehat{AKH}=\\widehat{ABH}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn $\\overset\\frown{AH}$) <br\/>Trong tam gi\u00e1c $ABC,$ $AD$ l\u00e0 \u0111\u01b0\u1eddng cao n\u00ean c\u0169ng l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $\\widehat{BAC}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow\\widehat{BAH}=\\widehat{CAH}$ <br\/> M\u1eb7t kh\u00e1c: X\u00e9t $\\Delta ABH$ v\u00e0 $\\Delta ACH$ c\u00f3: <br\/> $\\left\\{ \\begin{align} & AB=AC \\,(\\text{gi\u1ea3 thi\u1ebft}) \\\\ & \\widehat{BAH}=\\widehat{CAH}\\left( \\text{ch\u1ee9ng minh tr\u00ean} \\right) \\\\ & AH \\text{chung} \\\\ \\end{align} \\right.$ <br\/> $\\Rightarrow \\Delta ABH=\\Delta ACH\\,\\left( c.g.c \\right)$ <br\/> $\\Rightarrow \\widehat{ABH}=\\widehat{ACH}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow \\widehat{AKH}=\\widehat{ACH} $ <br\/> $\\Rightarrow \\Delta AKC$ c\u00e2n t\u1ea1i $A$ (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <br\/> $\\Rightarrow AK=AC$ (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c c\u00e2n) <\/span>"}]}],"id_ques":1496},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["4"]]],"list":[{"point":10,"width":50,"type_input":"","input_hint":["frac"],"ques":" <span class='basic_left'>Cho tam gi\u00e1c $ABC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O)$, $AC=2AB$ v\u00e0 $BC=6cm$. Ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $A$ c\u1eaft $BC$ \u1edf $M$. \u0110\u1ed9 d\u00e0i $AM$ l\u00e0 _input_ $(cm)$ ","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai15/lv3/img\/h933_K7.png' \/><\/center> <br\/><br\/> X\u00e9t $\\Delta MAB$ v\u00e0 $\\Delta MCA$ c\u00f3: <br\/> $\\widehat{M}$ chung <br\/> $\\widehat{MAB}=\\widehat{MCA}$ (g\u00f3c n\u1ed9i ti\u1ebfp v\u00e0 g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung c\u00f9ng ch\u1eafn $\\overset\\frown{AB}$) <br\/> $\\Rightarrow \\Delta MAB\\backsim \\Delta MCA(g.g)$ <br\/> $\\Rightarrow \\dfrac{MA}{MC}=\\dfrac{MB}{MA}=\\dfrac{AB}{AC}=\\dfrac{1}{2}$ (c\u00e1c c\u1eb7p t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng) <br\/> $\\Rightarrow \\left\\{ \\begin{align} & MA=\\dfrac{1}{2}MC \\\\ & M{{A}^{2}}=MB.MC \\\\ \\end{align} \\right.$ <br\/> $\\Rightarrow \\dfrac{1}{4}M{{C}^{2}}=MB.MC$ <br\/> $\\Rightarrow MB=\\dfrac{1}{4}MC\\Rightarrow MB=\\dfrac{1}{3}BC $ <br\/> $\\Rightarrow MB=2\\left( cm \\right)$ <br\/> $\\Rightarrow MC=4.MB=4.2=8\\left( cm \\right)$ <br\/> $\\Rightarrow MA = \\dfrac{1}{2}MC = \\dfrac{1}{2}.8 = 4 \\, (cm)$<br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $4$<\/span> <\/span>"}]}],"id_ques":1497},{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["EK","KE"]]],"list":[{"point":10,"width":50,"type_input":"","input_hint":["frac"],"ques":" <span class='basic_left'>Gi\u1ea3 s\u1eed $A$ v\u00e0 $B$ l\u00e0 hai \u0111i\u1ec3m ph\u00e2n bi\u1ec7t tr\u00ean \u0111\u01b0\u1eddng tr\u00f2n $(O)$. C\u00e1c ti\u1ebfp tuy\u1ebfn t\u1ea1i $A$ v\u00e0 $B$ c\u1eaft nhau t\u1ea1i $M$. T\u1eeb $A$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi $MB$, c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n $(O)$ t\u1ea1i $C$, $MC$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n $(O)$ t\u1ea1i $E$. C\u00e1c tia $AE$ v\u00e0 $MB$ c\u1eaft nhau t\u1ea1i $K$. Khi \u0111\u00f3, $MK^2=AK.$ _input_ ","hint":"D\u1ef1a v\u00e0o tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai15/lv3/img\/h933_K8.png' \/><\/center> <br\/><br\/> Ta c\u00f3: $MB\/\/AC$ <br\/> $\\Rightarrow \\widehat{BMC}=\\widehat{ACM}$ (hai g\u00f3c so le trong) <br\/> M\u1eb7t kh\u00e1c $\\widehat{ACM}=\\widehat{MAE}$ (g\u00f3c n\u1ed9i ti\u1ebfp v\u00e0 g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung c\u00f9ng ch\u1eafn $\\overset\\frown{AE}$) <br\/> $\\Rightarrow \\widehat{BMC}=\\widehat{MAE}$ <br\/> X\u00e9t $\\Delta MAK$ v\u00e0 $\\Delta EMK$c\u00f3: <br\/> $\\widehat{K}$ chung <br\/> $\\widehat{BMC}=\\widehat{MAE}$ (ch\u1ee9ng minh tr\u00ean) <br\/> $\\Rightarrow \\Delta MAK\\backsim \\Delta EMK\\,\\left( g.g \\right)$ <br\/> $\\Rightarrow \\dfrac{AK}{MK}=\\dfrac{MK}{EK}$ (c\u1eb7p t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng) <br\/> $\\Rightarrow M{{K}^{2}}=AK.EK$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0 $EK$<\/span> <\/span>"}]}],"id_ques":1498},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u (>; <; =) th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["="]]],"list":[{"point":10,"width":50,"type_input":"","ques":" <span class='basic_left'> Cho n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n $(O)$ \u0111\u01b0\u1eddng k\u00ednh $AB$, \u0111i\u1ec3m $C$ thu\u1ed9c n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n, $D$ l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a cung $AC$. K\u1ebb d\u00e2y $DE$ song song v\u1edbi $AB$. Ti\u1ebfp tuy\u1ebfn t\u1ea1i $E$ c\u1eaft $AB$ t\u1ea1i $K$. G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $DE$ v\u00e0 $AC$. So s\u00e1nh $DI$ v\u00e0 $BK$. <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $DI$ _input_$BK$.","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai15/lv3/img\/h933_K9.png' \/><\/center> <br\/> Ta c\u00f3: $AB\/\/DE$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow ABED$ l\u00e0 h\u00ecnh thang <br\/> $ABED$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O)$ n\u00ean $ABED$ l\u00e0 h\u00ecnh thang c\u00e2n <br\/> $\\overset\\frown{DA}=\\overset\\frown{BE}$ (hai cung b\u1ecb ch\u1eafn gi\u1eefa hai d\u00e2y song song) <br\/> $\\Rightarrow DA=BE$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa d\u00e2y v\u00e0 cung) <br\/> M\u00e0 $\\overset\\frown{DA}=\\overset\\frown{DC}$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow\\overset\\frown{DC}=\\overset\\frown{BE}$ <br\/> $\\Rightarrow \\widehat{DAC}=\\widehat{BEK}$ (g\u00f3c n\u1ed9i ti\u1ebfp v\u00e0 g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung ch\u1eafn hai cung b\u1eb1ng nhau) <br\/> Ta c\u00f3: $\\widehat{DEB}=\\widehat{EBK}$ (c\u00f9ng b\u00f9 $\\widehat{EBA}$) <br\/> $\\widehat{DEB}=\\widehat{ADI}$ (hai g\u00f3c k\u1ec1 c\u1ee7a h\u00ecnh thang c\u00e2n) <br\/> $\\Rightarrow \\widehat{EBK}=\\widehat{ADI}$ <br\/> X\u00e9t $\\Delta DAI$ v\u00e0 $\\Delta BEK$ c\u00f3: <br\/> $\\left\\{ \\begin{align} & \\widehat{DAI}=\\widehat{BEK} \\\\ & DA=BE \\\\ & \\widehat{ADI}=\\widehat{EBK} \\\\ \\end{align} \\right.$ (ch\u1ee9ng minh tr\u00ean) <br\/> $\\Rightarrow \\Delta DAI=\\Delta BEK\\,(g.c.g) $ <br\/> $\\Rightarrow DI=BK$ <br\/> <span class='basic_pink'> V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0 $=$ <\/span><\/span> "}]}],"id_ques":1499},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["226,3"]]],"list":[{"point":10,"width":50,"type_input":"","input_hint":["frac"],"ques":" <span class='basic_left'>M\u1ed9t ng\u01b0\u1eddi \u0111\u1ee9ng tr\u00ean m\u1ed9t \u0111\u1ec9nh n\u00fai thu\u1ed9c d\u00e3y Alpes cao $4000m$ th\u00ec nh\u00ecn th\u1ea5y c\u00f4ng tr\u00ecnh ki\u1ebfn tr\u00fac X. Bi\u1ebft b\u00e1n k\u00ednh tr\u00e1i \u0111\u1ea5t l\u00e0 $6400km$. Kho\u1ea3ng c\u00e1ch t\u1eeb \u0111i\u1ec3m nh\u00ecn c\u1ee7a ng\u01b0\u1eddi \u0111\u00f3 \u0111\u1ebfn c\u00f4ng tr\u00ecnh X t\u1ed1i \u0111a l\u00e0: _input_ $(km)$ <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai15/lv3/img\/h933_K10.1.png' \/><\/center> <br\/> (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t) ","explain":"<span class='basic_left'>Coi Tr\u00e1i \u0110\u1ea5t nh\u01b0 m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n l\u1edbn <br\/> T\u1eeb \u0111\u1ec9nh $M$ c\u1ee7a d\u00e3y Alpes ng\u01b0\u1eddi \u0111\u00f3 nh\u00ecn th\u1ea5y c\u00f4ng tr\u00ecnh ki\u1ebfn tr\u00fac A n\u00ean $MA$ \u0111\u01b0\u1ee3c coi nh\u01b0 l\u00e0 m\u1ed9t ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n, b\u00e1n k\u00ednh $R$ <br\/> T\u1eeb $M$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua t\u00e2m v\u00e0 c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n l\u1ea7n l\u01b0\u1ee3t t\u1ea1i $B$ v\u00e0 $C$ $\\Rightarrow BC= 2R$ <br\/> N\u1ed1i $C$ v\u1edbi $A$ ta \u0111\u01b0\u1ee3c h\u00ecnh v\u1ebd <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai15/lv3/img\/h933_K10.2.png' \/><\/center> <br\/><br\/> X\u00e9t $\\Delta MAB$ v\u00e0 $\\Delta MCA$ c\u00f3: <br\/> $\\widehat{M}$ chung <br\/> $\\widehat{MAB}=\\widehat{MCA}$ (g\u00f3c n\u1ed9i ti\u1ebfp v\u00e0 g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung c\u00f9ng ch\u1eafn $\\overset\\frown{AB}$) <br\/> $\\Rightarrow \\Delta MAB\\backsim \\Delta MCA(g.g)$ <br\/> $\\Rightarrow \\dfrac{MA}{MC}=\\dfrac{MB}{MA}$ (c\u1eb7p t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng) <br\/> $\\Rightarrow M{{A}^{2}}=MB.MC =MB.(MB+BC)$ <br\/> $= MB.(MB+2R) = 4.(4+6400.2) = 51216$ <br\/> $\\Rightarrow MA= 226,3 \\left( km \\right)$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $226,3$<\/span> <\/span>"}]}],"id_ques":1500}],"lesson":{"save":0,"level":3}}