{"segment":[{"time":24,"part":[{"time":3,"title":"Cho hai \u0111\u01b0\u1eddng tr\u00f2n $(O)$ v\u00e0 $(O\u2019)$ ti\u1ebfp x\u00fac ngo\u00e0i t\u1ea1i $A.$ \u0110\u01b0\u1eddng n\u1ed1i t\u00e2m c\u1eaft $(O)$ v\u00e0 $(O\u2019)$ t\u1ea1i \u0111i\u1ec3m th\u1ee9 hai t\u01b0\u01a1ng \u1ee9ng l\u00e0 $B$ v\u00e0 $C$. G\u1ecdi $EF$ l\u00e0 m\u1ed9t ti\u1ebfp tuy\u1ebfn chung ngo\u00e0i ($F$ thu\u1ed9c $(O)$ v\u00e0 $E$ thu\u1ed9c $(O\u2019)$). Ch\u1ee9ng minh r\u1eb1ng t\u1ee9 gi\u00e1c $BCEF$ n\u1ed9i ti\u1ebfp.","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[3],[4],[1],[6],[2],[5]]],"list":[{"point":10,"left":["$\\Rightarrow \\widehat{AFE}+\\widehat{AEF}=\\dfrac{1}{2}\\left( \\widehat{AOF}+\\widehat{AO'E} \\right)=\\dfrac{1}{2}{{.180}^{o}}={{90}^{o}}$ (1)"," $\\widehat{FBC}=\\widehat{EFA}$ (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 m\u1ed9t cung) (1) <br\/> $\\widehat{AEC}={{90}^{o}}$ (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n) (3) ","Ta c\u00f3: $OF\\bot EF;\\,O'E\\bot EF$ (t\u00ednh ch\u1ea5t ti\u1ebfp tuy\u1ebfn) <br\/> $\\Rightarrow OF\/\/O'E$ (t\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song) <br\/> $\\Rightarrow \\widehat{EO'O}+\\widehat{O'OF}$ (t\u1ed5ng hai g\u00f3c trong c\u00f9ng ph\u00eda)","$\\Rightarrow BCEF$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft)"," L\u1ea1i c\u00f3: $\\widehat{AFE}=\\dfrac{1}{2}\\widehat{AOF};\\,\\widehat{AEF}=\\dfrac{1}{2}\\widehat{AO'E}$ (g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung v\u00e0 g\u00f3c \u1edf t\u00e2m c\u00f9ng ch\u1eafn m\u1ed9t cung) "," T\u1eeb (1), (2) v\u00e0 (3) $\\Rightarrow \\widehat{FBC}+\\widehat{FEC}=\\widehat{EFA}+\\widehat{FEA}+\\widehat{AEC}={{90}^{o}}+{{90}^{o}}={{180}^{o}}$ "],"top":100,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai17/lv3/img\/h935_K1.png' \/><\/center> <br\/> Ta c\u00f3: $OF\\bot EF;\\,O'E\\bot EF$ (t\u00ednh ch\u1ea5t ti\u1ebfp tuy\u1ebfn) <br\/> $\\Rightarrow OF\/\/O'E$ (t\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song) <br\/> $\\Rightarrow \\widehat{EO'O}+\\widehat{O'OF}=180^o$ (t\u1ed5ng hai g\u00f3c trong c\u00f9ng ph\u00eda) <br\/> L\u1ea1i c\u00f3: $\\widehat{AFE}=\\dfrac{1}{2}\\widehat{AOF};\\,\\widehat{AEF}=\\dfrac{1}{2}\\widehat{AO'E}$ (g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung v\u00e0 g\u00f3c \u1edf t\u00e2m c\u00f9ng ch\u1eafn m\u1ed9t cung) <br\/> $\\Rightarrow \\widehat{AFE}+\\widehat{AEF}=\\dfrac{1}{2}\\left( \\widehat{AOF}+\\widehat{AO'E} \\right)=\\dfrac{1}{2}{{.180}^{o}}={{90}^{o}}$ (1) <br\/> $\\widehat{FBC}=\\widehat{EFA}$ (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 m\u1ed9t cung) (2) <br\/> $\\widehat{AEC}={{90}^{o}}$ (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n) (3) <br\/> T\u1eeb (1), (2) v\u00e0 (3) $\\Rightarrow \\widehat{FBC}+\\widehat{FEC}=\\widehat{EFA}+\\widehat{FEA}+\\widehat{AEC}={{90}^{o}}+{{90}^{o}}={{180}^{o}}$ <br\/> $\\Rightarrow BCEF$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <\/span>"}]}],"id_ques":1551},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'> Cho hai \u0111\u01b0\u1eddng tr\u00f2n $(O)$ v\u00e0 $(O\u2019)$ c\u1eaft nhau t\u1ea1i $A$ v\u00e0 $B$. Qua $A$ v\u1ebd hai c\u00e1t tuy\u1ebfn $CAD$ v\u00e0 $EAF$ ($C, E \\in (O); D, F \\in (O\u2019)$). \u0110\u01b0\u1eddng th\u1eb3ng $CE$ c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng $DF$ t\u1ea1i $P$. Ch\u1ee9ng minh t\u1ee9 gi\u00e1c $BEPF$ n\u1ed9i ti\u1ebfp.","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],[6],[1],[4],[2]]],"list":[{"point":10,"left":["M\u1eb7t kh\u00e1c $\\widehat{CAE}=\\widehat{DAF}$ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh) $\\Rightarrow \\widehat{CBE}=\\widehat{DAF}$"," M\u00e0 $ABFD$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O\u2019)$ <br\/> $\\Rightarrow \\widehat{BAD}+\\widehat{BFD}={{180}^{o}}$ (t\u1ed5ng hai g\u00f3c \u0111\u1ed1i nhau) hay $\\widehat{BEP}+\\widehat{BFP}={{180}^{o}}$ ","$\\Rightarrow $T\u1ee9 gi\u00e1c $BEPF$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft)","Ta c\u00f3: $ABEC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O)$ <br\/> $\\Rightarrow \\widehat{BCE}=\\widehat{BAF}$ (g\u00f3c ngo\u00e0i t\u1ea1i m\u1ed9t \u0111\u1ec9nh b\u1eb1ng g\u00f3c trong c\u1ee7a \u0111\u1ec9nh \u0111\u1ed1i di\u1ec7n)"," Ta c\u00f3: $\\widehat{BEP}=\\widehat{BCE}+\\widehat{CBE}$ (g\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c $BCE$) <br\/>$\\Rightarrow \\widehat{BEP}=\\widehat{BAF}+\\widehat{DAF}=\\widehat{BAD}$"," $\\widehat{CBE}=\\widehat{CAE}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn m\u1ed9t cung) "],"top":100,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai17/lv3/img\/h935_K2.png' \/><\/center> <br\/> Ta c\u00f3: $ABEC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O)$ <br\/> $\\Rightarrow \\widehat{BCE}=\\widehat{BAF}$ (g\u00f3c ngo\u00e0i t\u1ea1i m\u1ed9t \u0111\u1ec9nh b\u1eb1ng g\u00f3c trong c\u1ee7a \u0111\u1ec9nh \u0111\u1ed1i di\u1ec7n) <br\/> $\\widehat{CBE}=\\widehat{CAE}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn m\u1ed9t cung) <br\/> M\u1eb7t kh\u00e1c $\\widehat{CAE}=\\widehat{DAF}$ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh) <br\/> $\\Rightarrow \\widehat{CBE}=\\widehat{DAF}$ <br\/> Ta c\u00f3: $\\widehat{BEP}=\\widehat{BCE}+\\widehat{CBE}$ (g\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c $BCE$) <br\/> $\\Rightarrow \\widehat{BEP}=\\widehat{BAF}+\\widehat{DAF}=\\widehat{BAD}$ <br\/> M\u00e0 $ABFD$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O\u2019)$ <br\/> $\\Rightarrow \\widehat{BAD}+\\widehat{BFD}={{180}^{o}}$ (t\u1ed5ng hai g\u00f3c \u0111\u1ed1i nhau) hay $\\widehat{BEP}+\\widehat{BFP}={{180}^{o}}$ <br\/> $\\Rightarrow $T\u1ee9 gi\u00e1c $BEPF$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <\/span>"}]}],"id_ques":1552},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$, \u0111\u01b0\u1eddng cao $AH$. Tr\u00ean $AC$ l\u1ea5y \u0111i\u1ec3m $D$, $BD$ c\u1eaft $AH$ t\u1ea1i $M$. Qua $A$ v\u1ebd \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c $BD$ t\u1ea1i $N$ v\u00e0 c\u1eaft $BC$ t\u1ea1i $P$. Ch\u1ee9ng minh t\u1ee9 gi\u00e1c $NDCH$ n\u1ed9i ti\u1ebfp.","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[2],[6],[4],[1],[5],[3]]],"list":[{"point":10,"left":["$\\Rightarrow M$ l\u00e0 tr\u1ef1c t\u00e2m tam gi\u00e1c $\\Rightarrow PM\\bot AB$ "," $\\Rightarrow HNDC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) ","D\u1ec5 d\u00e0ng ch\u1ee9ng minh $MNPH$ n\u1ed9i ti\u1ebfp <br\/> $\\Rightarrow \\widehat{MNH}=\\widehat{MPH}\\,\\left( 2 \\right)$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn m\u1ed9t cung)","X\u00e9t $\\Delta ABP$ c\u00f3: $BN\\bot AP;\\,AH\\bot BP$ (gi\u1ea3 thi\u1ebft)"," T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\widehat{MNH}=\\widehat{HCD}$"," M\u00e0 $AC\\bot AB$ (gi\u1ea3 thi\u1ebft) $\\Rightarrow PM\/\/AC$ (t\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song) <br\/> $\\Rightarrow \\widehat{MPH}=\\widehat{HCD}\\,\\left( 1 \\right)$ (hai g\u00f3c \u0111\u1ed3ng v\u1ecb) "],"top":100,"hint":"S\u1eed d\u1ee5ng d\u1ea5u hi\u1ec7u g\u00f3c ngo\u00e0i t\u1ea1i m\u1ed9t \u0111\u1ec9nh b\u1eb1ng g\u00f3c trong c\u1ee7a \u0111\u1ec9nh \u0111\u1ed1i di\u1ec7n","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai17/lv3/img\/h935_K3.png' \/><\/center> <br\/> X\u00e9t $\\Delta ABP$ c\u00f3: <br\/> $BN\\bot AP;\\,AH\\bot BP$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow M$ l\u00e0 tr\u1ef1c t\u00e2m tam gi\u00e1c <br\/> $\\Rightarrow PM\\bot AB$ <br\/> M\u00e0 $\\,AC\\bot AB$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow PM\/\/AC$ (t\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song) <br\/> $\\Rightarrow \\widehat{MPH}=\\widehat{HCD}\\,\\left( 1 \\right)$ (hai g\u00f3c \u0111\u1ed3ng v\u1ecb) <br\/> D\u1ec5 d\u00e0ng ch\u1ee9ng minh $MNPH$ n\u1ed9i ti\u1ebfp <br\/> $\\Rightarrow \\widehat{MNH}=\\widehat{MPH}\\,\\left( 2 \\right)$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn m\u1ed9t cung) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\widehat{MNH}=\\widehat{HCD}$ <br\/> $\\Rightarrow HNDC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <\/span>"}]}],"id_ques":1553},{"time":24,"part":[{"time":3,"title":"Cho tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$, c\u00e1c trung tuy\u1ebfn $AH, BE, CF$ c\u1eaft nhau t\u1ea1i $G$. G\u1ecdi $M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BG$; $N$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $FG$. Ch\u1ee9ng minh r\u1eb1ng t\u1ee9 gi\u00e1c $CMNE$ n\u1ed9i ti\u1ebfp.","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],[5],[3],[6]]],"list":[{"point":10,"left":["$\\Rightarrow \\Delta BAG=\\Delta CAG\\,\\left( c.g.c \\right)$ <br\/> $\\Rightarrow\\widehat{ABG}=\\widehat{ACG}\\,(1)$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng)"," $\\Rightarrow MN\/\/BF$ $\\Rightarrow \\widehat{MNG}=\\widehat{ABG}\\,(2)$(hai g\u00f3c \u0111\u1ed3ng v\u1ecb) ","$\\Delta BAG$ v\u00e0 $\\Delta CAG$ c\u00f3: $\\left\\{ \\begin{align} & AB=AC\\,\\left( \\text{t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n} \\right) \\\\ & \\widehat{BAG}=\\widehat{CAG} \\,\\left( \\text{t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n} \\right)\\\\ & AG\\, \\text{chung} \\\\ \\end{align} \\right.$","T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\widehat{NME}=\\widehat{ECN}$ <br\/> $\\Rightarrow$ C\u00e1c \u0111i\u1ec3m $ M,\\,C$ nh\u00ecn $NE$ d\u01b0\u1edbi g\u00f3c $\\alpha $"," D\u1ec5 th\u1ea5y $MN$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a $\\Delta GBF$"," $\\Rightarrow MNEC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft)"],"top":100,"hint":"Ch\u1ee9ng minh hai \u0111\u1ec9nh c\u00f9ng nh\u00ecn m\u1ed9t c\u1ea1nh d\u01b0\u1edbi g\u00f3c kh\u00f4ng \u0111\u1ed5i","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai17/lv3/img\/h935_K4.png' \/><\/center> <br\/> X\u00e9t $\\Delta BAG$ v\u00e0 $\\Delta CAG$ c\u00f3: <br\/> $\\left\\{ \\begin{align} & AB=AC\\,\\left( \\text{t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n} \\right) \\\\ & \\widehat{BAG}=\\widehat{CAG} \\,\\left( \\text{t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n} \\right)\\\\ & AG\\, \\text{chung} \\\\ \\end{align} \\right.$ <br\/> $\\Rightarrow \\Delta BAG=\\Delta CAG\\,\\left( c.g.c \\right)$ <br\/> $\\Rightarrow\\widehat{ABG}=\\widehat{ACG}\\,(1)$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> D\u1ec5 th\u1ea5y $MN$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a $\\Delta GBF$ <br\/> $\\Rightarrow MN\/\/BF$ <br\/> $\\Rightarrow \\widehat{NMG}=\\widehat{ABG}\\,(2)$(hai g\u00f3c \u0111\u1ed3ng v\u1ecb) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\widehat{NME}=\\widehat{ECN}$ $\\Rightarrow$ C\u00e1c \u0111i\u1ec3m $ M,\\,C$ nh\u00ecn $NE$ d\u01b0\u1edbi g\u00f3c $\\alpha $ <br\/> $\\Rightarrow MNEC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <\/span>"}]}],"id_ques":1554},{"time":24,"part":[{"time":3,"title":"Cho tam gi\u00e1c $ABC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O)$. Tr\u00ean $(O)$ l\u1ea5y \u0111i\u1ec3m $M,$ g\u1ecdi $H,I,K$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 ch\u00e2n \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb $M$ xu\u1ed1ng c\u00e1c c\u1ea1nh $AB, AC, BC$. Ch\u1ee9ng minh r\u1eb1ng $H, I, K$ th\u1eb3ng h\u00e0ng.","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[3],[6],[2],[7],[4],[5],[1]]],"list":[{"point":10,"left":["$\\Rightarrow \\widehat{HAM}=\\widehat{HIM}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn cung $HM$) (2)"," T\u1eeb (1), (2) v\u00e0 (3) $\\Rightarrow \\widehat{MIK}+\\widehat{HIM}={{180}^{o}}$","T\u1ee9 gi\u00e1c $HAIM$ c\u00f3 $\\widehat{AHM}+\\widehat{AIM}={{90}^{o}}+{{90}^{o}}={{180}^{o}}$ n\u00ean n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n","$\\Rightarrow H,I,K$ th\u1eb3ng h\u00e0ng","T\u1ee9 gi\u00e1c $MIKC$ c\u00f3 $I, K$ c\u00f9ng nh\u00ecn $MC$ d\u01b0\u1edbi g\u00f3c $90^o$ n\u00ean n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n"," $\\Rightarrow \\widehat{MIK}+\\widehat{BCM}={{180}^{o}}$ (t\u1ed5ng hai g\u00f3c \u0111\u1ed1i nhau) (3)"," Ta c\u00f3 $ABCM$ n\u1ed9i ti\u1ebfp $(O)$ n\u00ean $\\widehat{BCM}=\\widehat{HAM}\\,(1)$ (g\u00f3c ngo\u00e0i t\u1ea1i m\u1ed9t \u0111\u1ec9nh b\u1eb1ng g\u00f3c trong c\u1ee7a \u0111\u1ec9nh \u0111\u1ed1i di\u1ec7n)"],"top":100,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai17/lv3/img\/h935_K5.png' \/><\/center> <br\/> Ta c\u00f3 $ABCM$ n\u1ed9i ti\u1ebfp $(O)$ <br\/> N\u00ean $\\widehat{BCM}=\\widehat{HAM}\\,(1)$ (g\u00f3c ngo\u00e0i t\u1ea1i m\u1ed9t \u0111\u1ec9nh b\u1eb1ng g\u00f3c trong c\u1ee7a \u0111\u1ec9nh \u0111\u1ed1i di\u1ec7n) <br\/> T\u1ee9 gi\u00e1c $HAIM$ c\u00f3 $\\widehat{AHM}+\\widehat{AIM}={{90}^{o}}+{{90}^{o}}={{180}^{o}}$ n\u00ean n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n <br\/> $\\Rightarrow \\widehat{HAM}=\\widehat{HIM}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn cung $HM$) (2) <br\/> T\u1ee9 gi\u00e1c $MIKC$ c\u00f3 $I, K$ c\u00f9ng nh\u00ecn $MC$ d\u01b0\u1edbi g\u00f3c $90^o$ n\u00ean n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n <br\/> $\\Rightarrow \\widehat{MIK}+\\widehat{BCM}={{180}^{o}}$ (t\u1ed5ng hai g\u00f3c \u0111\u1ed1i nhau) (3) <br\/> T\u1eeb (1), (2) v\u00e0 (3) $\\Rightarrow \\widehat{MIK}+\\widehat{HIM}={{180}^{o}}$ <br\/> $\\Rightarrow H,I,K$ th\u1eb3ng h\u00e0ng <\/span>"}]}],"id_ques":1555},{"time":24,"part":[{"time":3,"title":"Cho h\u00ecnh thang $ABCD$ c\u00f3 \u0111\u00e1y l\u1edbn $AD$, \u0111\u00e1y nh\u1ecf $BC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O)$. G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AB$ v\u00e0 $DC$, ti\u1ebfp tuy\u1ebfn t\u1ea1i $B$ v\u00e0 $D$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n c\u1eaft nhau t\u1ea1i $K$. Ch\u1ee9ng minh t\u1ee9 gi\u00e1c $BIKD$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n.","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[4],[5],[1],[3],[7],[6],[2]]],"list":[{"point":10,"left":["$\\widehat{DKB}=\\dfrac{\\text{s\u0111}\\overset\\frown{DAB}-\\text{s\u0111}\\overset\\frown{DCB}}{2}=\\dfrac{\\text{s\u0111}\\overset\\frown{DA}+\\text{s\u0111}\\overset\\frown{AB}-\\text{s\u0111}\\overset\\frown{DC}-\\text{s\u0111}\\overset\\frown{BC}}{2}=\\dfrac{\\text{s\u0111}\\overset\\frown{DA}-\\text{s\u0111}\\overset\\frown{BC}}{2}\\,$"," $\\Rightarrow \\widehat{DIB}=\\widehat{DKB}$","Ta c\u00f3 h\u00ecnh thang $ABCD$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n n\u00ean $ABCD$ l\u00e0 h\u00ecnh thang c\u00e2n"," $\\widehat{DIB}=\\dfrac{\\text{s\u0111}\\overset\\frown{DA}-\\text{s\u0111}\\overset\\frown{BC}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m \u1edf ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n)","$\\Rightarrow BIKD$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) "," $\\Rightarrow$ C\u00e1c \u0111i\u1ec3m $ I,K$ c\u00f9ng nh\u00ecn $BD$ d\u01b0\u1edbi m\u1ed9t g\u00f3c $\\alpha $"," $\\Rightarrow AB=CD$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow \\overset\\frown{AB}=\\overset\\frown{CD}$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y)"],"top":75,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai17/lv3/img\/h935_K6.png' \/><\/center> <br\/> Ta c\u00f3 h\u00ecnh thang $ABCD$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n n\u00ean $ABCD$ l\u00e0 h\u00ecnh thang c\u00e2n <br\/> $\\Rightarrow AB=CD$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow \\overset\\frown{AB}=\\overset\\frown{CD}$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y) <br\/> $\\widehat{DIB}=\\dfrac{\\text{s\u0111}\\overset\\frown{DA}-\\text{s\u0111}\\overset\\frown{BC}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m \u1edf ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\widehat{DKB}=\\dfrac{\\text{s\u0111}\\overset\\frown{DAB}-\\text{s\u0111}\\overset\\frown{DCB}}{2}=\\dfrac{\\text{s\u0111}\\overset\\frown{DA}+\\text{s\u0111}\\overset\\frown{AB}-\\text{s\u0111}\\overset\\frown{DC}-\\text{s\u0111}\\overset\\frown{BC}}{2}=\\dfrac{\\text{s\u0111}\\overset\\frown{DA}-\\text{s\u0111}\\overset\\frown{BC}}{2}\\,$ <br\/> $\\Rightarrow \\widehat{DIB}=\\widehat{DKB}$ <br\/> $\\Rightarrow$ C\u00e1c \u0111i\u1ec3m $ I,K$ c\u00f9ng nh\u00ecn $BD$ d\u01b0\u1edbi m\u1ed9t g\u00f3c $\\alpha $ <br\/> $\\Rightarrow BIKD$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <\/span>"}]}],"id_ques":1556},{"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":[[["120"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" <span class='basic_left'>Cho tam gi\u00e1c $ABC$ c\u00f3 g\u00f3c $B$ b\u1eb1ng $60^o$. Hai tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $A$ v\u00e0 $C$ c\u1eaft nhau \u1edf $I$, c\u1eaft $BC$ \u1edf $D$, c\u1eaft $AB$ \u1edf $E$. S\u1ed1 \u0111o g\u00f3c $EID$ l\u00e0 _input_ $^o$ ","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai17/lv3/img\/h935_K7.png' \/><\/center> <br\/>Ta c\u00f3 $\\widehat{AIC}+\\widehat{IAC}+\\widehat{ICA}={{180}^{o}}$ (t\u1ed5ng ba g\u00f3c c\u1ee7a tam gi\u00e1c $IAC$) <br\/> $\\widehat{ABC}+\\widehat{ACB}+\\widehat{BAC}={{180}^{o}}$ (t\u1ed5ng ba g\u00f3c c\u1ee7a tam gi\u00e1c $ABC$) <br\/> $\\Rightarrow \\widehat{AIC}={{180}^{o}}-\\left( \\widehat{IAC}+\\widehat{ICA} \\right)={{180}^{o}}-\\dfrac{\\widehat{BAC}+\\widehat{ACB}}{2}\\\\ ={{180}^{o}}-\\dfrac{{{180}^{o}}-\\widehat{ABC}}{2}\\\\ ={{180}^{o}}-{{60}^{o}}={{120}^{o}}$ <br\/> $\\Rightarrow \\widehat{EID}=\\widehat{AIC}={{120}^{o}}$ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh) <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $120$<\/span> <\/span>"}]}],"id_ques":1557},{"time":24,"part":[{"time":3,"title":"Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$ d\u00e2y cung $AB$. C\u00e1c ti\u1ebfp tuy\u1ebfn v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $A$ v\u00e0 $B$ c\u1eaft nhau t\u1ea1i $C$. Tr\u00ean d\u00e2y $AB$ l\u1ea5y \u0111i\u1ec3m $E$ ($EA > EB$). \u0110\u01b0\u1eddng vu\u00f4ng g\u00f3c v\u1edbi $OE$ t\u1ea1i $E$ c\u1eaft $CA$ v\u00e0 $CB$ theo th\u1ee9 t\u1ef1 \u1edf $I$ v\u00e0 $K$. Ch\u1ee9ng minh $OICK$ n\u1ed9i ti\u1ebfp.","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],[5],[7],[6],[2],[4]]],"list":[{"point":10,"left":["$\\Rightarrow \\widehat{AIO}=\\widehat{AEO}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn m\u1ed9t cung) (1)"," Ta c\u00f3: $\\widehat{OAC}=\\widehat{OBC}={{90}^{o}}$ (t\u00ednh ch\u1ea5t ti\u1ebfp tuy\u1ebfn)","$\\Rightarrow \\widehat{AEO}=\\widehat{OKC}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn m\u1ed9t cung) (2)"," $\\Rightarrow IOKC$ n\u1ed9i ti\u1ebfp trong \u0111\u01b0\u1eddng tr\u00f2n (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft)","T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\widehat{AIO}=\\widehat{OKC}$ "," T\u1ee9 gi\u00e1c $AOEI$ c\u00f3 $A, E$ nh\u00ecn $IO$ d\u01b0\u1edbi g\u00f3c $90^o$ n\u00ean n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n"," T\u1ee9 gi\u00e1c $OEBK$ c\u00f3 $B, E$ nh\u00ecn $OK$ d\u01b0\u1edbi g\u00f3c $90^o$ n\u00ean n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n"],"top":55,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai17/lv3/img\/h935_K8.png' \/><\/center> <br\/> Ta c\u00f3: $\\widehat{OAC}=\\widehat{OBC}={{90}^{o}}$ (t\u00ednh ch\u1ea5t ti\u1ebfp tuy\u1ebfn) <br\/> T\u1ee9 gi\u00e1c $AOEI$ c\u00f3 $A, E$ nh\u00ecn $IO$ d\u01b0\u1edbi g\u00f3c $90^o$ n\u00ean n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n <br\/> $\\Rightarrow \\widehat{AIO}=\\widehat{AEO}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn m\u1ed9t cung) (1) <br\/> T\u1ee9 gi\u00e1c $OEBK$ c\u00f3 $B, E$ nh\u00ecn $OK$ d\u01b0\u1edbi g\u00f3c $90^o$ n\u00ean n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n <br\/> $\\Rightarrow \\widehat{AEO}=\\widehat{OKC}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn m\u1ed9t cung) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\widehat{AIO}=\\widehat{OKC}$ <br\/> $\\Rightarrow IOKC$ n\u1ed9i ti\u1ebfp trong \u0111\u01b0\u1eddng tr\u00f2n (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <\/span>"}]}],"id_ques":1558},{"time":24,"part":[{"time":3,"title":"Cho tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$, \u0111\u01b0\u1eddng cao $AD$, tr\u1ef1c t\u00e2m $H$. G\u1ecdi $E$ l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi $H$ qua $D$. Ch\u1ee9ng minh $ABEC$ l\u00e0 t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp.","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[2],[1],[4],[3],[5]]],"list":[{"point":10,"left":["$\\Rightarrow BHCE$ l\u00e0 h\u00ecnh thoi (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) $\\Rightarrow \\left\\{ \\begin{align} & BH\/\/EC \\\\ & BE\/\/HC \\\\ \\end{align} \\right.$ (t\u00ednh ch\u1ea5t h\u00ecnh thoi)"," X\u00e9t t\u1ee9 gi\u00e1c $BHCE$ c\u00f3: $\\left\\{ \\begin{align} & HE\\bot BC;HD=HE\\,\\left( \\text{t\u00ednh ch\u1ea5t \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng} \\right) \\\\ & BD=DC\\,\\left( \\text{t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n} \\right) \\\\ \\end{align} \\right.$","$\\Rightarrow$ C\u00e1c \u0111i\u1ec3m $ B,C$ c\u00f9ng nh\u00ecn $AE$ d\u01b0\u1edbi g\u00f3c $90^o$ "," M\u00e0 $\\left\\{ \\begin{align} & BH\\bot AC \\\\ & CH\\bot AB \\\\ \\end{align} \\right.$ $\\Rightarrow \\left\\{ \\begin{align} & EC\\bot AC \\\\ & EB\\bot AB \\\\ \\end{align} \\right.$ $\\Rightarrow \\left\\{ \\begin{align} & \\widehat{ACE}={{90}^{o}} \\\\ & \\widehat{ABE}={{90}^{o}} \\\\ \\end{align} \\right.$ "," $\\Rightarrow ABEC$ l\u00e0 t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft)"],"top":100,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai17/lv3/img\/h935_K9.png' \/><\/center> <br\/> X\u00e9t t\u1ee9 gi\u00e1c $BHCE$ c\u00f3: <br\/> $\\left\\{ \\begin{align} & HE\\bot BC;HD=HE\\,\\left( \\text{t\u00ednh ch\u1ea5t \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng} \\right) \\\\ & BD=DC\\,\\left( \\text{t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n} \\right) \\\\ \\end{align} \\right.$ <br\/> $\\Rightarrow BHCE$ l\u00e0 h\u00ecnh thoi (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <br\/> $\\Rightarrow \\left\\{ \\begin{align} & BH\/\/EC \\\\ & BE\/\/HC \\\\ \\end{align} \\right.$ (t\u00ednh ch\u1ea5t h\u00ecnh thoi) <br\/> M\u00e0 $\\left\\{ \\begin{align} & BH\\bot AC \\\\ & CH\\bot AB \\\\ \\end{align} \\right.$ $\\Rightarrow \\left\\{ \\begin{align} & EC\\bot AC \\\\ & EB\\bot AB \\\\ \\end{align} \\right.$ $\\Rightarrow \\left\\{ \\begin{align} & \\widehat{ACE}={{90}^{o}} \\\\ & \\widehat{ABE}={{90}^{o}} \\\\ \\end{align} \\right.$ <br\/> $\\Rightarrow$ C\u00e1c \u0111i\u1ec3m $ B,C$ c\u00f9ng nh\u00ecn $AE$ d\u01b0\u1edbi g\u00f3c $90^o$ <br\/> $\\Rightarrow ABEC$ l\u00e0 t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <\/span>"}]}],"id_ques":1559},{"time":24,"part":[{"time":3,"title":"Cho tam gi\u00e1c nh\u1ecdn $ABC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O)$, \u0111\u01b0\u1eddng k\u00ednh $AK$. G\u1ecdi $E$ v\u00e0 $F$ theo th\u1ee9 t\u1ef1 l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a $B$ v\u00e0 $C$ tr\u00ean $AK$, $H$ l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a $A$ tr\u00ean $BC$. Ch\u1ee9ng minh r\u1eb1ng $\\widehat{EHF}=\\widehat{BAC}$.","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[4],[1],[3],[6],[5],[2]]],"list":[{"point":10,"left":["$\\Rightarrow \\widehat{HFA}=\\widehat{HCA}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn m\u1ed9t cung) (2)"," T\u1ee9 gi\u00e1c $ABHE$ c\u00f3 $H, E$ nh\u00ecn $AB$ d\u01b0\u1edbi g\u00f3c $90^o$ n\u00ean n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n","T\u1ee9 gi\u00e1c $ACFH$ c\u00f3 $H, F$ nh\u00ecn $AC$ d\u01b0\u1edbi g\u00f3c $90^o$ n\u00ean n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n "," T\u1eeb (1), (2) v\u00e0 (3) $\\Rightarrow \\widehat{BAC}=\\widehat{EHF}$"," M\u00e0 $\\widehat{ABH}+\\widehat{HCA}+\\widehat{BAC}=\\widehat{HEF}+\\widehat{HFA}+\\widehat{EHF}={{180}^{o}}$ (t\u1ed5ng ba g\u00f3c c\u1ee7a tam gi\u00e1c $ABC$ v\u00e0 $EHF$) (3)"," $\\Rightarrow \\widehat{ABH}=\\widehat{HEF}$ (g\u00f3c ngo\u00e0i t\u1ea1i m\u1ed9t \u0111\u1ec9nh b\u1eb1ng g\u00f3c trong c\u1ee7a \u0111\u1ec9nh \u0111\u1ed1i di\u1ec7n) (1)"],"top":75,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai17/lv3/img\/h935_K10.png' \/><\/center> <br\/> T\u1ee9 gi\u00e1c $ABHE$ c\u00f3 $H, E$ nh\u00ecn $AB$ d\u01b0\u1edbi g\u00f3c $90^o$ n\u00ean n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n <br\/> $\\Rightarrow \\widehat{ABH}=\\widehat{HEF}$ (g\u00f3c ngo\u00e0i t\u1ea1i m\u1ed9t \u0111\u1ec9nh b\u1eb1ng g\u00f3c trong c\u1ee7a \u0111\u1ec9nh \u0111\u1ed1i di\u1ec7n) (1) <br\/> T\u1ee9 gi\u00e1c $ACFH$ c\u00f3 $H, F$ nh\u00ecn $AC$ d\u01b0\u1edbi g\u00f3c $90^o$ n\u00ean n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n <br\/> $\\Rightarrow \\widehat{HFA}=\\widehat{HCA}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn m\u1ed9t cung) (2) <br\/> M\u00e0 $\\widehat{ABH}+\\widehat{HCA}+\\widehat{BAC}=\\widehat{HEF}+\\widehat{HFA}+\\widehat{EHF}={{180}^{o}}$ (t\u1ed5ng ba g\u00f3c c\u1ee7a tam gi\u00e1c $ABC$ v\u00e0 $EHF$) (3) <br\/> T\u1eeb (1), (2) v\u00e0 (3) $\\Rightarrow \\widehat{BAC}=\\widehat{EHF}$ <\/span>"}]}],"id_ques":1560}],"lesson":{"save":0,"level":3}}