{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":" $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$, \u0111\u01b0\u1eddng cao $AH$. L\u1ea5y \u0111i\u1ec3m $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AC$, $K$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AH$, $D$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $HC$. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $I$ l\u00e0 giao \u0111i\u1ec3m ba \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $\\triangle{AHC}$ ","B. $KD \/\/ AC$","C. $BK \\perp AD$ ","D. C\u1ea3 ba \u0111\u00e1p \u00e1n tr\u00ean \u0111\u1ec1u \u0111\u00fang"],"hint":"","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv3/img\/H7C3B23_K1.png' \/><\/center><br\/> $\\blacktriangleright$ $\\triangle{AHC}$ vu\u00f4ng t\u1ea1i $H$ c\u00f3 $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AC$ <br\/> $\\Rightarrow$ $HI$ l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn $\\Rightarrow$ $HI = AI = IC$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) <br\/> $IA = IH$ $\\Rightarrow$ $I$ n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AH$ (1) <br\/> $IH = IC$ $\\Rightarrow$ $I$ n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $HC$ (2) <br\/> $IA = IC$ $\\Rightarrow$ $I$ n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AC$ (3) <br\/> T\u1eeb (1), (2), (3) $\\Rightarrow$ $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a ba \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $\\triangle{AHC}$ <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n A \u0111\u00fang <\/b> <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{KHD}$ v\u00e0 $\\triangle{DIK}$ c\u00f3: <br\/> $KD$ chung <br\/> M\u00e0 $AH \\perp BC; DI \\perp BC \\Rightarrow AH \/\/ DI$ $\\Rightarrow$ $\\widehat{HDK} = \\widehat{KDI}$ (so le trong) <br\/> $IK \\perp AH; BC \\perp AH$ $\\Rightarrow$ $KI \/\/ BC$ $\\Rightarrow$ $\\widehat{HKD} = \\widehat{IKD}$ (so le trong) <br\/> V\u1eady $\\triangle{KHD} = \\triangle{DIK}$ (g.c.g) <br\/> $\\Rightarrow$ $HK = ID; HD = KI$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> X\u00e9t hai tam gi\u00e1c vu\u00f4ng $KHD$ v\u00e0 $DIC$ c\u00f3: <br\/> $HK = ID$ (cmt) <br\/> $HD = HC$ ($ID$ l\u00e0 trung tr\u1ef1c) <br\/> $\\Rightarrow$ $\\triangle{KHD} = \\triangle{DIC}$ (hai c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow$ $\\widehat{HDK} = \\widehat{DCI}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow$ $KD \/\/ AC$ (v\u00ec hai g\u00f3c \u0111\u1ed3ng v\u1ecb b\u1eb1ng nhau) <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n B \u0111\u00fang <\/b> <br\/> $\\blacktriangleright$ Ta c\u00f3 $KD \/\/ AC$ (cmt) v\u00e0 $AB \\perp AC$ <br\/> $\\Rightarrow$ $KD \\perp AB$ <br\/> X\u00e9t $\\triangle{ABD}$ c\u00f3: $AH \\perp BD$ (gi\u1ea3 thi\u1ebft) v\u00e0 $KD \\perp AB$ <br\/> N\u00ean $K$ l\u00e0 tr\u1ef1c t\u00e2m c\u1ee7a $\\triangle{ABD}$ <br\/> V\u1eady $KB$ l\u00e0 \u0111\u01b0\u1eddng cao th\u1ee9 ba c\u1ee7a $\\triangle{ABD}$ suy ra $BK \\perp AD$ <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n C \u0111\u00fang <\/b> <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t l\u00e0: D <\/span> ","column":2}]}],"id_ques":2001},{"time":24,"part":[{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["1","2","3"]],"list":[{"point":10,"img":"","ques":"Cho tam gi\u00e1c $ABC$, \u0111\u01b0\u1eddng cao $AH$. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a $AH$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $AD = BC$. T\u1ea1i $B$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng $BE \\perp AB$ ($E$ v\u00e0 $C$ thu\u1ed9c hai n\u1eeda m\u1eb7t ph\u1eb3ng \u0111\u1ed1i nhau b\u1edd l\u00e0 $AB$). T\u1ea1i $C$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng $CF \\perp AC$ v\u00e0 $CF = AC$ ($F$ v\u00e0 $B$ thu\u1ed9c hai n\u1eeda m\u1eb7t ph\u1eb3ng \u0111\u1ed1i nhau b\u1edd $AC$).","hint":"","column":2,"number_true":2,"select":["A. $DC = BF$","B. $DC \\perp BF$","C. Ba \u0111\u01b0\u1eddng th\u1eb3ng $DH, BF, CE$ \u0111\u1ed3ng quy","D. $DC > BF$ "],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv3/img\/H7C3B23_K2.png' \/><\/center> <br\/> G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $CD$ v\u00e0 $BF$ v\u00e0 $G$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BD$ v\u00e0 $EC$ <br\/> $\\blacktriangleright$ $\\triangle{DAC}$ v\u00e0 $\\triangle{BCF}$ c\u00f3: <br\/> $DA = DC$ (gt) <br\/> $\\widehat{DAC} = \\widehat{BCF}$ (g\u00f3c c\u00f3 c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng vu\u00f4ng g\u00f3c, c\u00f9ng t\u00f9 $CB \\perp AD, CF \\perp AC$) <br\/> $AC = CF$ (gt) <br\/> $\\Rightarrow$ $\\triangle{DAC} = \\triangle{BCF}$ (c.g.c) <br\/> $\\Rightarrow$ $DC = BF$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <b> \u0110\u00e1p \u00e1n A \u0111\u00fang, D sai <\/b> <br\/> $\\blacktriangleright$ $\\widehat{C_{1}} = \\widehat{F}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> M\u00e0 $\\widehat{C_{1}} + \\widehat{C_{2}} = 90^{o}$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow$ $\\widehat{F} + \\widehat{C_{2}} = 90^{o}$ <br\/> Trong $\\triangle{CFI}$ c\u00f3: $\\widehat{F} + \\widehat{C_{2}} = 90^{o}$ $\\Rightarrow$ $\\widehat{CIF} = 90^{o}$ <br\/> V\u1eady $DC \\perp BF$ <b> (\u0110\u00e1p \u00e1n B \u0111\u00fang) <\/b> <br\/> $\\blacktriangleright$ Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta \u0111\u01b0\u1ee3c $\\triangle{DAB} = \\triangle{CBE}$ (c.g.c) <br\/> $\\Rightarrow$ $\\widehat{B_{1}} = \\widehat{E}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> M\u00e0 $\\widehat{B_{1}} + \\widehat{B_{2}} = 90^{o}$ $\\Rightarrow$ $\\widehat{E} + \\widehat{B_{2}} = 90^{o}$ <br\/> Trong $\\triangle{EBG}$ c\u00f3: $\\widehat{E} + \\widehat{B_{2}} = 90^{o}$ $\\Rightarrow$ $\\widehat{EBG} = 90^{o}$ <br\/> $\\Rightarrow$ $DB \\perp CE$ <br\/> Trong $\\triangle{DBC}$ c\u00f3 $DH \\perp BC; BI \\perp AC; CG \\perp AB$ <br\/> V\u1eady $DH; BI; CG$ l\u00e0 ba \u0111\u01b0\u1eddng cao c\u1ee7a $\\triangle{DBC}$ hay ba \u0111\u01b0\u1eddng cao \u0111\u1ed3ng quy (t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c) <br\/> <b> \u0110\u00e1p \u00e1n C \u0111\u00fang <\/b> <br\/> <span class='basic_pink'> V\u1eady c\u00e1c \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A, B, C<\/span> "}]}],"id_ques":2002},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" Cho \u0111o\u1ea1n th\u1eb3ng $AB$. Tr\u00ean c\u00f9ng m\u1ed9t n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $AB$, v\u1ebd hai tia $Ax, By$ c\u00f9ng vu\u00f4ng g\u00f3c v\u1edbi $AB$. Tr\u00ean hai tia $Ax, By$ l\u1ea7n l\u01b0\u1ee3t l\u1ea5y hai \u0111i\u1ec3m $C, D$ sao cho $AC = \\dfrac{1}{2}BD$. V\u1ebd $BE$ vu\u00f4ng g\u00f3c v\u1edbi $AD$ $(E \\in AD)$ v\u00e0 g\u1ecdi $F$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $ED$. Khi \u0111\u00f3 $CF \\perp BF$. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG ","B. SAI"],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> G\u1ecdi $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $EB$, ch\u1ee9ng minh $AH \\perp BF$ d\u1ef1a v\u00e0o t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng cao trong tam gi\u00e1c $ABF$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $CF \/\/ AH$ b\u1eb1ng c\u00e1ch ch\u1ee9ng minh hai g\u00f3c \u1edf v\u1ecb tr\u00ed so le trong b\u1eb1ng nhau <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u1eeb b\u01b0\u1edbc 1 v\u00e0 b\u01b0\u1edbc 2 suy ra \u0111i\u1ec1u c\u1ea7n ch\u1ee9ng minh <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv3/img\/H7C3B23_K3.png' \/><\/center><br\/> $\\blacktriangleright$ G\u1ecdi $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BE$. Khi \u0111\u00f3 $HF$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a $\\triangle{BDE}$ \u1ee9ng v\u1edbi \u0111\u00e1y $BD$ (xem b\u00e0i $64$ s\u00e1ch b\u00e0i t\u1eadp To\u00e1n $7$ t\u1eadp $1$ trang $146$) <br\/> $\\Rightarrow$ $HF \/\/ BD$ v\u00e0 $HF = \\dfrac{1}{2}BD$ (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng trung b\u00ecnh) <br\/> M\u00e0 $BD \\perp AB$ (gt) n\u00ean suy ra $HF \\perp AB$ <br\/> V\u00ec $BE \\perp AD$ (gt) v\u00e0 $BE, HF$ c\u1eaft nhau t\u1ea1i $H$ <br\/> $\\Rightarrow$ $H$ l\u00e0 tr\u1ef1c t\u00e2m tam gi\u00e1c $ABF$ (t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng cao trong tam gi\u00e1c) <br\/> $\\Rightarrow$ $AH \\perp BF$ (t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng cao trong tam gi\u00e1c) (1) <br\/> $\\blacktriangleright$ T\u1eeb $HF = \\dfrac{1}{2}BD $ (cmt) v\u00e0 $AC = \\dfrac{1}{2}BD$ (gt) $\\Rightarrow$ $HF = AC$ <br\/> V\u00ec $HF \/\/ BD$ v\u00e0 $AC \/\/ BD$(v\u00ec c\u00f9ng vu\u00f4ng g\u00f3c v\u1edbi $AB$) <br\/> $\\Rightarrow$ $HF \/\/ AC$ (t\u00ednh ch\u1ea5t) <br\/> $\\Rightarrow$ $\\widehat{CAF} = \\widehat{HFA}$ (hai g\u00f3c so le trong) <br\/> Do \u0111\u00f3, $\\triangle{ACF} = \\triangle{FHA}$ (c.g.c) <br\/> $\\Rightarrow$ $\\widehat{CFA} = \\widehat{HAF}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng). <br\/> M\u00e0 hai g\u00f3c n\u00e0y \u1edf v\u1ecb tr\u00ed so le n\u00ean $CF \/\/ AH$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $BF \\perp CF$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span> ","column":2}]}],"id_ques":2003},{"time":24,"part":[{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["1","2","3"]],"list":[{"point":10,"img":"","ques":"Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$, \u0111\u01b0\u1eddng cao $AH$. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a c\u1ee7a tia $AH$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $AD = AH$. G\u1ecdi $E$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $HC$, $F$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $DE$ v\u00e0 $AC$. G\u1ecdi $P$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AH$. H\u00e3y ch\u1ecdn nh\u1eefng kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang.","hint":"","column":1,"number_true":3,"select":["A. $H, F$ v\u00e0 trung \u0111i\u1ec3m $M$ c\u1ee7a $DC$ th\u1eb3ng h\u00e0ng","B. $HF = \\dfrac{1}{3}DC$","C. $EP \\perp AB$ ","D. $BP \\perp DC$, $CP$ kh\u00f4ng vu\u00f4ng g\u00f3c v\u1edbi $DB$ "],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv3/img\/H7C3B23_K5.png' \/><\/center> <br\/>$\\blacktriangleright$ X\u00e9t $\\triangle{CDH}$ c\u00f3 $CA$ v\u00e0 $DE$ l\u00e0 hai trung tuy\u1ebfn <br\/> $\\Rightarrow$ $F$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a $\\triangle{CDH}$ <br\/> $\\Rightarrow$ \u0110\u01b0\u1eddng trung tuy\u1ebfn $HF$ \u0111i qua trung \u0111i\u1ec3m $M$ c\u1ee7a $CD$ hay ba \u0111i\u1ec3m $H, F, M$ th\u1eb3ng h\u00e0ng <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n A \u0111\u00fang <\/b> <br\/> $\\blacktriangleright$ V\u00ec $F$ l\u00e0 tr\u1ecdng t\u00e2m $\\triangle{CDH}$ (cmt) <br\/> $\\Rightarrow$ $HF = \\dfrac{2}{3}HM$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) (1) <br\/> M\u1eb7t kh\u00e1c, v\u00ec $HM$ l\u00e0 trung tuy\u1ebfn \u1ee9ng v\u1edbi c\u1ea1nh huy\u1ec1n c\u1ee7a $\\triangle{CDH}$ vu\u00f4ng t\u1ea1i $H$ n\u00ean $HM = \\dfrac{1}{2}CD$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $HF = \\dfrac{2}{3}HM = \\dfrac{2}{3}.\\dfrac{1}{2}CD = \\dfrac{1}{3}CD$ <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n B \u0111\u00fang <\/b> <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ACH}$ c\u00f3 $E$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $CH$, $P$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AH$ <br\/> $\\Rightarrow$ $EP$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh \u1ee9ng v\u1edbi \u0111\u00e1y $AC$ (xem b\u00e0i $64$ s\u00e1ch b\u00e0i t\u1eadp To\u00e1n $7$ t\u1eadp $1$ trang $146$) <br\/> $\\Rightarrow$ $EP \/\/ AC$ m\u00e0 $AB \\perp AC$ $\\Rightarrow$ $EP \\perp AB$ (t\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song) <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n C \u0111\u00fang <\/b> <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABE}$, ta c\u00f3 $AH$ v\u00e0 $EP$ l\u00e0 hai \u0111\u01b0\u1eddng cao <br\/> $\\Rightarrow$ $P$ l\u00e0 tr\u1ef1c t\u00e2m c\u1ee7a $\\triangle{ABE}$ $\\Rightarrow$ $BP \\perp AE$ (t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng cao trong tam gi\u00e1c) (3) <br\/> M\u1eb7t kh\u00e1c, x\u00e9t $\\triangle{CDH}$ c\u00f3 $E$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $CH$, $A$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $DH$ <br\/> $\\Rightarrow$ $AE$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh \u1ee9ng v\u1edbi \u0111\u00e1y $CD$ <br\/> $\\Rightarrow$ $AE \/\/ CD$ (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng trung b\u00ecnh) (4) <br\/> T\u1eeb (3) v\u00e0 (4) $\\Rightarrow$ $BP \\perp CD$ (t\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song) <br\/> X\u00e9t $\\triangle{BCD}$, ta c\u00f3 $DH$ v\u00e0 $BP$ l\u00e0 hai \u0111\u01b0\u1eddng cao <br\/> $\\Rightarrow$ $P$ l\u00e0 tr\u1ef1c t\u00e2m $\\triangle{BCD}$ <br\/> $\\Rightarrow$ $CP \\perp BD$ (t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c) <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n D sai <\/b> <br\/> <span class='basic_pink'> V\u1eady c\u00e1c \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A, B, C<\/span> "}]}],"id_ques":2004},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["t","f","t","f"]],"list":[{"point":10,"image":"","col_name":["Cho tam gi\u00e1c $ABC$ nh\u1ecdn v\u1edbi tr\u1ef1c t\u00e2m $H$. <br\/> Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00e2y \u0111\u00fang hay sai?","\u0110\u00fang","Sai"],"arr_ques":[" $HA + HB + HC < AB + AC$"," $HA + HB + HC = AB + AC$"," $HA + HB + HC < \\dfrac{2}{3}(AB + AC + BC)$ "," $HA + HB + HC > \\dfrac{2}{3}(AB + AC + BC)$ "],"hint":"","explain":["<span class='basic_left'> \u0110\u00daNG v\u00ec: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv3/img\/H7C3B23_K4.png' \/><\/center> <br\/> $\\blacktriangleright$ Qua $H$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi $AC$ c\u1eaft $AB$ t\u1ea1i $E$ v\u00e0 k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi $AB$ c\u1eaft $AC$ t\u1ea1i $F$ <br\/> Khi \u0111\u00f3 $HF = AE$ (t\u00ednh ch\u1ea5t \u0111o\u1ea1n ch\u1eafn song song) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{HAF}$ c\u00f3: $HA < AF + HF = AF + AE$ (b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c) (1) <br\/> $\\blacktriangleright$ V\u00ec $HE \/\/ AC$ v\u00e0 $BH \\perp AC$ $\\Rightarrow$ $HE \\perp BH$ <br\/> Do \u0111\u00f3 ta c\u00f3 $HB < EB$ (quan h\u1ec7 gi\u1eefa \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c, \u0111\u01b0\u1eddng xi\u00ean) (2) <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u0169ng c\u00f3 $HC < FC$ (3) <br\/> $\\blacktriangleright$ C\u1ed9ng theo v\u1ebf c\u1ee7a (1), (2), (3) ta \u0111\u01b0\u1ee3c: <br\/> $HA + HB + HC < AE + AF + EB + FC = AB + AC$ <\/span> ","<br\/> <span class='basic_left'> SAI: Theo ch\u1ee9ng minh tr\u00ean <\/span> ","<br\/> <span class='basic_left'> \u0110\u00daNG v\u00ec: <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 c\u00e2u \u0111\u1ea7u, ta c\u00f3: <br\/> $HA + HB + HC < AB + AC$ <br\/> $HA + HB + HC < BA + BC$ <br\/> $HA + HB + HC < CA + CB$ <br\/> C\u1ed9ng theo t\u1eebng v\u1ebf c\u1ee7a c\u00e1c b\u1ea5t \u0111\u1eb3ng th\u1ee9c tr\u00ean ta \u0111\u01b0\u1ee3c: <br\/> $3(HA + HB + HC) < 2(AB + AC + BC)$ <br\/> $\\Rightarrow$ $HA + HB + HC < \\dfrac{2}{3}(AB + AC + BC)$ <\/span>","<span class='basic_left'> SAI: Theo ch\u1ee9ng minh tr\u00ean <\/span> "]}]}],"id_ques":2005},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 $AB < AC$, \u0111\u01b0\u1eddng cao $AH$. Tr\u00ean tia $HA$ l\u1ea5y \u0111i\u1ec3m $P$ sao cho $HP = HB$ v\u00e0 tr\u00ean tia $HC$ l\u1ea5y \u0111i\u1ec3m $M$ sao cho $HM = HA$. Khi \u0111\u00f3 $HA < HB < HC$ v\u00e0 $P$ l\u00e0 kh\u00f4ng ph\u1ea3i l\u00e0 tr\u1ef1c t\u00e2m tam gi\u00e1c $ABM$. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG ","B. SAI"],"hint":"","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv3/img\/H7C3B23_K6.png' \/><\/center><br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABC}$, v\u00ec $AB < AC$ (gt) n\u00ean $\\widehat{ABC} > \\widehat{C}$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n) <br\/> M\u00e0 $\\widehat{C} = \\widehat{BAH}$ (c\u00f9ng ph\u1ee5 v\u1edbi $\\widehat{ABC}$) <br\/> $\\Rightarrow$ $\\widehat{ABC} > \\widehat{BAH}$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{BAH}$, c\u00f3: $\\widehat{ABC} > \\widehat{ABH}$ (cmt) <br\/> $\\Rightarrow$ $HB < HA$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong tam gi\u00e1c) (1) <br\/> $\\blacktriangleright$ L\u1ea1i c\u00f3 $\\widehat{ABC} = \\widehat{CAH}$ (c\u00f9ng ph\u1ee5 v\u1edbi g\u00f3c $C$) <br\/> $\\Rightarrow$ $\\widehat{CAH} > \\widehat{C} $ <br\/> $\\Rightarrow$ $HA < HC$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong $\\triangle{ACH}$) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $HB < HA < HC$ <br\/> $\\Rightarrow$ Kh\u1eb3ng \u0111\u1ecbnh $HA < HB < HC$ l\u00e0 <b> SAI <\/b> <br\/> $\\blacktriangleright$ V\u00ec $HP = HB$ (gt) m\u00e0 $HB < HA$ (cmt) <br\/> $\\Rightarrow$ $HP < HA$ $\\Rightarrow$ $P$ n\u1eb1m tr\u00ean \u0111o\u1ea1n $HA$ <br\/> $\\triangle{HBP}$ vu\u00f4ng c\u00e2n t\u1ea1i $H$ (v\u00ec $HB = HP$) $\\Rightarrow$ $\\widehat{B_{1}} = 45^{o}$ <br\/> T\u01b0\u01a1ng t\u1ef1 $\\triangle{AHM}$ vu\u00f4ng c\u00e2n t\u1ea1i $H$ (v\u00ec $AH = HM$) $\\Rightarrow$ $\\widehat{M_{1}} = 45^{o}$ <br\/> G\u1ecdi $N$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BP$ v\u00e0 $AM$ <br\/> Khi \u0111\u00f3 $\\triangle{MNB}$ vu\u00f4ng c\u00e2n t\u1ea1i $N$ (v\u00ec $\\widehat{M_{1}} = \\widehat{B_{1}} = 45^{o}$) <br\/> $\\Rightarrow$ $BP \\perp AM$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABM}$, c\u00f3 $AH \\perp BM$ (gt) v\u00e0 $BP \\perp AM$ (cmt) <br\/> $\\Rightarrow$ $P$ l\u00e0 tr\u1ef1c t\u00e2m c\u1ee7a $\\triangle{ABM}$ (t\u00ednh ch\u1ea5t) <br\/> $\\Rightarrow$ Kh\u1eb3ng \u0111\u1ecbnh $P$ kh\u00f4ng ph\u1ea3i tr\u1ef1c t\u00e2m $\\triangle{ABM}$ l\u00e0 <b> SAI <\/b> <br\/> V\u1eady \u0111\u00e1p \u00e1n ph\u1ea3i ch\u1ecdn l\u00e0 <span class='basic_pink'> SAI <\/span> ","column":2}]}],"id_ques":2006},{"time":24,"part":[{"title":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u","title_trans":"Cho tam gi\u00e1c nh\u1ecdn $ABC$, hai \u0111\u01b0\u1eddng cao $BD$ v\u00e0 $CE$. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $BD$ l\u1ea5y \u0111i\u1ec3m $I$ sao cho $BI = AC$, tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $CE$ l\u1ea5y \u0111i\u1ec3m $K$ sao cho $CK = AB$. H\u00e3y s\u1eafp x\u1ebfp c\u00e1c c\u00e2u sau \u0111\u1ec3 ch\u1ee9ng minh tam gi\u00e1c $AIK$ vu\u00f4ng c\u00e2n. ","temp":"sequence","correct":[[[2],[1],[5],[3],[4]]],"list":[{"point":10,"image":"","left":[" $\\triangle{ABI} = \\triangle{KCA}$ (c.g.c) v\u00ec: $\\begin{cases} BI = AC (gt) \\\\ \\widehat{ABI} = \\widehat{ACK} (cmt) \\\\ AB = CK (gt) \\end{cases}$ ","Ta c\u00f3: $\\widehat{ABD} = \\widehat{ACE}$ (c\u00f9ng ph\u1ee5 v\u1edbi $\\widehat{BAC}$) <br\/> $\\Rightarrow$ $\\widehat{ABI} = \\widehat{ACK}$ (L\u1ea7n l\u01b0\u1ee3t k\u1ec1 b\u00f9 v\u1edbi $\\widehat{ABD}$ v\u00e0 $\\widehat{ACE}$) "," T\u1eeb (2) v\u00e0 (3) $\\Rightarrow$ $\\widehat{IAD} + \\widehat{KAC} = 90^{o}$ hay $\\widehat{IAK} = 90^{o}$ (4) <br\/> T\u1eeb (1) v\u00e0 (4) $\\Rightarrow$ $\\triangle{IAK}$ vu\u00f4ng c\u00e2n t\u1ea1i $A$ "," $\\Rightarrow$ $AI = AK$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> V\u00e0 $\\widehat{KAC} = \\widehat{AIB}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) (2)"," $\\triangle{AID}$ vu\u00f4ng \u1edf $D$ c\u00f3 $\\widehat{IAD} + \\widehat{AID} = 90^{o}$ (3) "],"top":100,"hint":"","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv3/img\/H7C3B23_K7.png' \/><\/center> $\\blacktriangleright$ Ta c\u00f3: $\\widehat{ABD} = \\widehat{ACE}$ (c\u00f9ng ph\u1ee5 v\u1edbi $\\widehat{BAC}$) <br\/> $\\Rightarrow$ $\\widehat{ABI} = \\widehat{ACK}$ (L\u1ea7n l\u01b0\u1ee3t k\u1ec1 b\u00f9 v\u1edbi $\\widehat{ABD}$ v\u00e0 $\\widehat{ACE}$) <br\/> $\\triangle{ABI} = \\triangle{KCA}$ (c.g.c) v\u00ec: $\\begin{cases} BI = AC (gt) \\\\ \\widehat{ABI} = \\widehat{ACK} (cmt) \\\\ AB = CK (gt) \\end{cases}$ <br\/> $\\Rightarrow$ $AI = AK$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> V\u00e0 $\\widehat{KAC} = \\widehat{AIB}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) (2) <br\/> $\\triangle{AID}$ vu\u00f4ng \u1edf $D$ c\u00f3 $\\widehat{IAD} + \\widehat{AID} = 90^{o}$ (3) <br\/> T\u1eeb (2) v\u00e0 (3) $\\Rightarrow$ $\\widehat{IAD} + \\widehat{KAC} = 90^{o}$ hay $\\widehat{IAK} = 90^{o}$ (4) <br\/> T\u1eeb (1) v\u00e0 (4) $\\Rightarrow$ $\\triangle{IAK}$ vu\u00f4ng c\u00e2n t\u1ea1i $A$ <\/span> "}]}],"id_ques":2007},{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["5"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 c\u00e1c \u0111\u01b0\u1eddng cao $BE, CF$ c\u1eaft nhau t\u1ea1i $H$. G\u1ecdi $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n $AH$, $K$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh $BC$. T\u00ednh $IK$ bi\u1ebft $AH = 6cm, BC = 8cm$ <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $IK = $ _input_ $cm$ ","hint":"Ch\u1ee9ng minh $\\triangle{IFK}$ vu\u00f4ng r\u1ed3i t\u00ednh $IK$ d\u1ef1a v\u00e0o $\\triangle{IFK}$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $\\widehat{IAF} = \\widehat{IFA}$ v\u00e0 $\\widehat{KFB} = \\widehat{KBF}$ r\u1ed3i t\u00ednh $\\widehat{IFK}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $IK$ d\u1ef1a v\u00e0o \u0111\u1ecbnh l\u00fd Pitago <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv3/img\/H7C3B23_K8.png' \/><\/center> <br\/> Tam gi\u00e1c $ABC$ c\u00f3 c\u00e1c \u0111\u01b0\u1eddng cao $BE, CF$ c\u1eaft nhau t\u1ea1i $H$ <br\/> $\\Rightarrow$ $H$ l\u00e0 tr\u1ecdng t\u00e2m $\\triangle{ABC}$, g\u1ecdi $D$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AH$ v\u1edbi $BC$ <br\/> $\\Rightarrow$ $AH \\perp BC$ t\u1ea1i $D$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{AFH}$ vu\u00f4ng \u1edf $F$, c\u00f3 trung tuy\u1ebfn $FI$ <br\/> $\\Rightarrow$ $FI = \\dfrac{1}{2}AH = IA$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) <br\/> Do \u0111\u00f3 $\\triangle{FAI}$ c\u00e2n t\u1ea1i $I$ (v\u00ec $AI = FI$) <br\/> $\\Rightarrow$ $\\widehat{IFA} = \\widehat{IAF}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) (1) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{BFC}$ vu\u00f4ng t\u1ea1i $F$, c\u00f3 trung tuy\u1ebfn $FK$ <br\/> $\\Rightarrow$ $FK = \\dfrac{1}{2}BC = BK$ <br\/> Do \u0111\u00f3 $\\triangle{FBK}$ c\u00e2n t\u1ea1i $K$ (v\u00ec $FK = BK$) <br\/> $\\Rightarrow$ $\\widehat{KFB} = \\widehat{KBF}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $\\widehat{IFA} + \\widehat{KFB} = \\widehat{IAF} + \\widehat{KBF} = 90^{o}$ (v\u00ec $\\triangle{ADB}$ vu\u00f4ng t\u1ea1i $D$) <br\/> $\\Rightarrow$ $\\widehat{IFK} = 90^{o}$ hay $\\triangle{IFK}$ vu\u00f4ng t\u1ea1i $F$ <br\/> $\\blacktriangleright$ Theo ch\u1ee9ng minh tr\u00ean $FI = \\dfrac{1}{2}AH = \\dfrac{1}{2}.6cm = 3cm$ <br\/> $FK = \\dfrac{1}{2}BC = \\dfrac{1}{2}.8 = 4cm$ <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pitago cho tam gi\u00e1c $IFK$ vu\u00f4ng t\u1ea1i $F$, ta c\u00f3: <br\/> $IK^2 = FI^2 + FK^2 = 3^2 + 4^2 = 25$ <br\/> $\\Rightarrow$ $IK = 5cm$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $5$ <\/span>"}]}],"id_ques":2008},{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["45"],["135"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 tr\u1ef1c t\u00e2m $H$ ($H$ kh\u00e1c $A$). Bi\u1ebft $AH = BC$. T\u00ednh s\u1ed1 \u0111\u00f3 g\u00f3c $BAC$ <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $\\widehat{BAC} = $ _input_ $^o$ ho\u1eb7c _input_$^o$ ","hint":"X\u00e9t hai tr\u01b0\u1eddng h\u1ee3p $\\widehat{BAC} < 90^o$ v\u00e0 $\\widehat{BAC} > 90^{o}$ ","explain":"<span class='basic_left'> $\\blacktriangleright$ Ta th\u1ea5y $\\widehat{BAC} \\neq 90^{o}$ v\u00ec $H \\neq A$ <br\/> G\u1ecdi $K$ l\u00e0 ch\u00e2n \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb $B$ xu\u1ed1ng $AC$ <br\/> <b> Tr\u01b0\u1eddng h\u1ee3p 1: <\/b> $\\widehat{BAC} < 90^{o}$ <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv3/img\/H7C3B23_K91.png' \/><\/center><br\/> X\u00e9t hai tam gi\u00e1c vu\u00f4ng $AKH $ v\u00e0 $BKC$, c\u00f3: <br\/> C\u1ea1nh huy\u1ec1n: $AH = BC$ (gt) <br\/> G\u00f3c nh\u1ecdn: $\\widehat{HAK} = \\widehat{KBC}$ ($= 90^{o} - \\widehat{C}$) <br\/> $\\Rightarrow$ $\\triangle{AKH} = \\triangle{BKC}$ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow$ $AK = BK$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow$ $\\triangle{AKB}$ vu\u00f4ng c\u00e2n t\u1ea1i $K$ <br\/> $\\Rightarrow$ $\\widehat{BAC} = 45^{o}$ <br\/> <b> Tr\u01b0\u1eddng h\u1ee3p 2:<\/b> $\\widehat{BAC} > 90^{o}$ <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv3/img\/H7C3B23_K92.png' \/><\/center> <br\/> <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 tr\u01b0\u1eddng h\u1ee3p 1 ta \u0111\u01b0\u1ee3c: <br\/> $HK = CK$ $\\Rightarrow$ $\\triangle{HKC}$ vu\u00f4ng c\u00e2n t\u1ea1i $K$ $\\Rightarrow$ $\\widehat{CHK} = 45^{o}$ <br\/> $A$ l\u00e0 tr\u1ef1c t\u00e2m $\\triangle{HBC}$ ($CK \\perp BH, HA \\perp BC$) n\u00ean $BA \\perp HC$ <br\/> $\\Rightarrow$ Hai g\u00f3c $BAC$ v\u00e0 $BHC$ c\u00f3 c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng vu\u00f4ng g\u00f3c <br\/> V\u00e0 $\\widehat{BAC} > 90^{o}$ n\u00ean $\\widehat{BAC} + \\widehat{BHC} = 180^{o}$ (t\u00ednh ch\u1ea5t hai g\u00f3c c\u00f3 c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng vu\u00f4ng g\u00f3c) <br\/> $\\Rightarrow$ $\\widehat{BAC} = 180^{o} - \\widehat{BHC} = 180^{o} - 45^{o} = 135^{o}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $45$; $135$ <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: N\u1ebfu g\u00f3c c\u00f3 c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng vu\u00f4ng g\u00f3c: <br\/> Ch\u00fang b\u1eb1ng nhau n\u1ebfu c\u1ea3 hai g\u00f3c c\u00f9ng nh\u1ecdn ho\u1eb7c c\u00f9ng t\u00f9 <br\/> Ch\u00fang b\u00f9 nhau n\u1ebfu g\u00f3c n\u00e0y nh\u1ecdn g\u00f3c kia t\u00f9 <\/i><\/span> "}]}],"id_ques":2009},{"time":24,"part":[{"title":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u","title_trans":"Cho tam gi\u00e1c $ABC$ \u0111\u1ec1u, c\u1ea1nh $a$ v\u00e0 $M$ l\u00e0 \u0111i\u1ec3m b\u1ea5t k\u00ec n\u1eb1m trong tam gi\u00e1c. H\u00e3y s\u1eafp x\u1ebfp c\u00e1c \u00fd ch\u1ee9ng minh r\u1eb1ng t\u1ed5ng kho\u1ea3ng c\u00e1ch t\u1eeb $M$ \u0111\u1ebfn ba c\u1ea1nh c\u1ee7a tam gi\u00e1c $ABC$ kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o v\u1ecb tr\u00ed \u0111i\u1ec3m $M$ ","temp":"sequence","correct":[[[2],[4],[1],[3]]],"list":[{"point":10,"image":"","left":[" K\u1ebb \u0111\u01b0\u1eddng cao $AH$ c\u1ee7a $\\triangle{ABC}$, ta c\u00f3 di\u1ec7n t\u00edch $\\triangle{ABC}$ l\u00e0: <br\/> $S_{\\triangle{ABC}} = \\dfrac{1}{2}AH.BC = \\dfrac{1}{2}AH.a$ (1) ","T\u1eeb (1) v\u00e0 (2) suy ra $MI + MI + MK = AH$ (kh\u00f4ng \u0111\u1ed5i) <br\/> V\u1eady gi\u00e1 tr\u1ecb c\u1ee7a d kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o v\u1ecb tr\u00ed \u0111i\u1ec3m $M$ "," G\u1ecdi $I, J, K$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a $M$ tr\u00ean $BC, CA, AB$ <br\/> Khi \u0111\u00f3 t\u1ed5ng kho\u1ea3ng c\u00e1ch t\u1eeb $M$ \u0111\u1ebfn ba c\u1ea1nh c\u1ee7a tam gi\u00e1c $ABC$ l\u00e0: $d = MI + MJ + MK$ "," $\\begin{align} \\text{M\u1eb7t} \\hspace{0,2cm} \\text{kh\u00e1c} \\hspace{0,2cm} S_{\\triangle{ABC}} &= S_{\\triangle{BMC}} + S_{\\triangle{AMB}} + S_{\\triangle{CMA}} \\\\ & = \\dfrac{1}{2}MI.BC + \\dfrac{1}{2}MK.AB + \\dfrac{1}{2}MJ.AC \\\\ &= \\dfrac{1}{2}a(MI + MJ + MK) (2) \\end{align} $ "],"top":120,"hint":"","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv3/img\/H7C3B23_K10.png' \/><\/center> G\u1ecdi $I, J, K$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a $M$ tr\u00ean $BC, CA, AB$ <br\/> Khi \u0111\u00f3 t\u1ed5ng kho\u1ea3ng c\u00e1ch t\u1eeb $M$ \u0111\u1ebfn ba c\u1ea1nh c\u1ee7a tam gi\u00e1c $ABC$ l\u00e0: $d = MI + MJ + MK$ <br\/> K\u1ebb \u0111\u01b0\u1eddng cao $AH$ c\u1ee7a $\\triangle{ABC}$, ta c\u00f3 di\u1ec7n t\u00edch $\\triangle{ABC}$ l\u00e0: <br\/> $S_{\\triangle{ABC}} = \\dfrac{1}{2}AH.BC = \\dfrac{1}{2}AH.a$ (1) <br\/> M\u1eb7t kh\u00e1c ta l\u1ea1i c\u00f3: <br\/> $\\begin{align} S_{\\triangle{ABC}} &= S_{\\triangle{BMC}} + S_{\\triangle{AMB}} + S_{\\triangle{CMA}} \\\\ &= \\dfrac{1}{2}MI.BC + \\dfrac{1}{2}MK.AB + \\dfrac{1}{2}MJ.AC \\\\ &= \\dfrac{1}{2}a(MI + MJ + MK) (2) \\end{align} $ <br\/> T\u1eeb (1) v\u00e0 (2) suy ra $MI + MJ + MK = AH$ (kh\u00f4ng \u0111\u1ed5i) <br\/> V\u1eady gi\u00e1 tr\u1ecb c\u1ee7a d kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o v\u1ecb tr\u00ed \u0111i\u1ec3m $M$ <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: Trong b\u00e0i gi\u1ea3i tr\u00ean ta \u0111\u00e3 s\u1eed d\u1ee5ng di\u1ec7n t\u00edch nh\u01b0 m\u1ed9t c\u1ea7u n\u1ed1i trung gian gi\u1eefa hai \u0111\u1ea1i \u0111\u01b0\u1ee3ng h\u00ecnh h\u1ecdc. Ph\u01b0\u01a1ng ph\u00e1p n\u00e0y c\u00f3 t\u00ean g\u1ecdi l\u00e0 \"Ph\u01b0\u01a1ng ph\u00e1p di\u1ec7n t\u00edch\" <\/i> <\/span> "}]}],"id_ques":2010}],"lesson":{"save":0,"level":3}}