{"segment":[{"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,"content":"","type_input":"","ques":"Cho tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $BA$ l\u1ea5y \u0111i\u1ec3m $D$ v\u00e0 tr\u00ean c\u1ea1nh $AC$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $BD = CE$. H\u00e3y so s\u00e1nh $BC$ v\u00e0 $DE$. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $BC$ _input_ $DE$ ","hint":"S\u1eed d\u1ee5ng quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong tam gi\u00e1c","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai24/lv3/img\/H7C3B24_K1.png' \/><\/center><br\/> $\\blacktriangleright$ $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$ $\\Rightarrow$ $\\widehat{ABC} = \\widehat{ACB}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> X\u00e9t $\\triangle{BCD}$, c\u00f3: $\\widehat{ABC} > \\widehat{BDC}$ (t\u00ednh ch\u1ea5t g\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c) <br\/> M\u1eb7t kh\u00e1c: $\\widehat{DCE} > \\widehat{ACB} = \\widehat{ABC}$ <br\/> T\u1eeb \u0111\u00f3 suy ra $\\widehat{DCE} > \\widehat{BDC}$ (t\u00ednh ch\u1ea5t b\u1eafc c\u1ea7u) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{BCD}$ v\u00e0 $\\triangle{CDE}$, c\u00f3: <br\/> $\\begin{cases} BD = CE \\hspace{0,2cm} (gt) \\\\ \\widehat{BDC} < \\widehat{DCE} \\hspace{0,2cm} (cmt) \\\\ CD \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $BC < DE$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong hai tam gi\u00e1c) <br\/><span class='basic_pink'> V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0: $<$ <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: T\u1eeb b\u00e0i to\u00e1n n\u00e0y ta suy ra k\u1ebft qu\u1ea3 sau: <br\/> Trong tam gi\u00e1c $ABC$ c\u00f3 g\u00f3c $A$ v\u00e0 $AB + AC$ kh\u00f4ng \u0111\u1ed5i, tam gi\u00e1c c\u00e2n (t\u1ea1i $A$) l\u00e0 tam gi\u00e1c c\u00f3 chu vi nh\u1ecf nh\u1ea5t <br\/> Quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong hai tam gi\u00e1c: N\u1ebfu hai tam gi\u00e1c c\u00f3 hai c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng b\u1eb1ng nhau nh\u01b0ng c\u00e1c g\u00f3c xen gi\u1eefa kh\u00f4ng b\u1eb1ng nhau th\u00ec \u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c l\u1edbn h\u01a1n l\u00e0 c\u1ea1nh l\u1edbn h\u01a1n. \u0110\u1ea3o l\u1ea1i \u0111\u1ed1i di\u1ec7n v\u1edbi c\u1ea1nh l\u1edbn h\u01a1n l\u00e0 g\u00f3c l\u1edbn h\u01a1n <\/i><\/span> "}]}],"id_ques":2031},{"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_random","correct":[[["<"],["="]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Cho $\\widehat{xOy} = 60^{o}$, $A$ l\u00e0 \u0111i\u1ec3m tr\u00ean tia $Ox$, $B$ l\u00e0 \u0111i\u1ec3m tr\u00ean tia $Oy$ ($A, B$ kh\u00f4ng tr\u00f9ng v\u1edbi $O$). H\u00e3y so s\u00e1nh $OA + OB$ v\u1edbi $2AB$ <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $OA + OB$ _input_ $2AB$ ho\u1eb7c $OA + OB$ _input_ $2AB$ ","hint":"S\u1eed d\u1ee5ng quan h\u1ec7 gi\u1eefa \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c - \u0111\u01b0\u1eddng xi\u00ean - h\u00ecnh chi\u1ebfu <br\/> V\u1ebd tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $xOy$","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai24/lv3/img\/H7C3B24_K2.png' \/><\/center><br\/> $\\blacktriangleright$ K\u1ebb tia ph\u00e2n gi\u00e1c $Ot$ c\u1ee7a $\\widehat{xOy}$. <br\/> G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AB$ v\u00e0 $Ot$; $H, K$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a $A, B$ tr\u00ean $Ot$ <br\/> $\\blacktriangleright$ V\u00ec $Ot$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{xOy}$ n\u00ean $\\widehat{xOt} = \\widehat{yOt} = \\dfrac{\\widehat{xOy}}{2} = \\dfrac{60^{o}}{2} = 30^{o}$ <br\/> X\u00e9t $\\triangle{OAH}$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: <br\/> $\\widehat{AOH} = 30^{o}$ n\u00ean $OA = 2AH$ (trong tam gi\u00e1c vu\u00f4ng c\u1ea1nh \u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c $30^{o}$ b\u1eb1ng n\u1eeda c\u1ea1nh huy\u1ec1n) <br\/> V\u00ec $AH, AI$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c, \u0111\u01b0\u1eddng xi\u00ean k\u1ebb t\u1eeb $A$ \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng $Ot$ n\u00ean $AH \\leq AI$ <br\/> Do v\u1eady: $OA \\leq 2AI$ (1) <br\/> $\\blacktriangleright$ Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u00f3 <br\/> $OB = 2BK \\leq 2BI$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) ta c\u00f3: $OA + OB \\leq 2AI + 2BI = 2(AI + BI) = 2AB$ <br\/> D\u1ea5u \"=\" x\u1ea3y ra khi $H \\equiv I \\equiv K$ hay $AB \\perp Ot$ <br\/><span class='basic_pink'> V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0: $<$ v\u00e0 $=$ <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: Ch\u00eca kh\u00f3a \u0111\u1ec3 gi\u1ea3i b\u00e0i to\u00e1n tr\u00ean l\u00e0 vi\u1ec7c v\u1ebd tia ph\u00e2n gi\u00e1c $Ot$. Nh\u1eefng b\u01b0\u1edbc ti\u1ebfp theo tr\u1edf n\u00ean \u0111\u01a1n gi\u1ea3n h\u01a1n v\u1edbi vi\u1ec7c s\u1eed d\u1ee5ng t\u00ednh ch\u1ea5t c\u1ee7a tam gi\u00e1c vu\u00f4ng c\u00f3 g\u00f3c $30^o$ v\u00e0 quan h\u1ec7 gi\u1eefa \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c v\u00e0 \u0111\u01b0\u1eddng xi\u00ean <\/i><\/span> "}]}],"id_ques":2032},{"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,"content":"","type_input":"","ques":"Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$. G\u1ecdi $H$ l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a \u0111i\u1ec3m $A$ tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng $BC$. H\u00e3y so s\u00e1nh $AH + BC$ v\u00e0 $AB + AC$ <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $AH + BC$ _input_ $AB + AC$ ","hint":"S\u1eed d\u1ee5ng quan h\u1ec7 gi\u1eefa \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c - \u0111\u01b0\u1eddng xi\u00ean","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai24/lv3/img\/H7C3B24_K3.png' \/><\/center><br\/> $\\blacktriangleright$ K\u1ebb \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c $AH$ xu\u1ed1ng $BC$ $\\Rightarrow$ $H \\in BC$ <br\/> Tr\u00ean tia $CB$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $CE = CA$, tr\u00ean tia $AB$ l\u1ea5y \u0111i\u1ec3m $F$ sao cho $AF = AH$ <br\/> $\\blacktriangleright$ Ta c\u00f3: $AH < CA$ (quan h\u1ec7 \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c - \u0111\u01b0\u1eddng xi\u00ean) <br\/> $\\Rightarrow$ $CH < CE$ (v\u00ec $CE = CA$) <br\/> M\u1eb7t kh\u00e1c: $CA < CB$ (quan h\u1ec7 \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c - \u0111\u01b0\u1eddng xi\u00ean) <br\/> $\\Rightarrow$ $CE < CB$ (v\u00ec $CE = CA$) <br\/> T\u1eeb \u0111\u00f3 ta c\u00f3: $CH < CE < CB$ hay $E$ n\u1eb1m gi\u1eefa $H$ v\u00e0 $B$ <br\/> V\u00ec $AH < AB$ (quan h\u1ec7 \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c - \u0111\u01b0\u1eddng xi\u00ean) <br\/> $\\Rightarrow$ $AF < AB$ $\\Rightarrow$ $F$ n\u1eb1m gi\u1eefa $A$ v\u00e0 $B$ <br\/> $\\blacktriangleright$ V\u00ec $CE = CA$ $\\Rightarrow$ $\\triangle{ACE}$ c\u00e2n t\u1ea1i $C$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $\\Rightarrow$ $\\widehat{CAE} = \\widehat{E_{1}}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> M\u00e0 $\\widehat{CAE} + \\widehat{A_{1}} = 90^{o}$ <br\/> $\\widehat{E_{1}} + \\widehat{A_{2}} = 90^{o}$ ($\\triangle{AHE}$ vu\u00f4ng t\u1ea1i $E$) <br\/> $\\Rightarrow$ $\\widehat{A_{1}} = \\widehat{A_{2}}$ <br\/> Do \u0111\u00f3 d\u1ec5 d\u00e0ng ch\u1ec9 ra \u0111\u01b0\u1ee3c $\\triangle{AHE} = \\triangle{AFE}$ (c.g.c) <br\/> $\\Rightarrow$ $\\widehat{AFE} = \\widehat{AHE} = 90^{o}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> Hay $EF \\perp AB$ <br\/> $\\blacktriangleright$ Ta c\u00f3: $AH + BC = AF + CE + BE$ <br\/> $AB + AC = AF + BF + CE$ <br\/> M\u00e0 $BE > BF$ (quan h\u1ec7 \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c - \u0111\u01b0\u1eddng xi\u00ean) <br\/> $\\Rightarrow$ $AH + BC > AB + AC$ <br\/><span class='basic_pink'> V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0: $>$ <\/span> "}]}],"id_ques":2033},{"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 tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{B} = 75^{o}, \\widehat{C} = 60^{o}$. \u0110i\u1ec3m $M$ n\u1eb1m trong tam gi\u00e1c $ABC$ sao cho $\\triangle{MBC}$ vu\u00f4ng c\u00e2n t\u1ea1i $M$. Khi \u0111\u00f3 $MA = MB$. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>?","select":["A. \u0110\u00daNG ","B. SAI"],"hint":"Ch\u1ee9ng minh b\u1eb1ng ph\u1ea3n ch\u1ee9ng, gi\u1ea3 s\u1eed $MA \\neq MB$ v\u00e0 x\u00e9t hai tr\u01b0\u1eddng h\u1ee3p $MA > MB$ v\u00e0 $MA < MB$","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai24/lv3/img\/H7C3B24_K4.png' \/><\/center><br\/> $\\blacktriangleright$ $\\triangle{ABC}$ c\u00f3: $\\widehat{ABC} + \\widehat{ACB} + \\widehat{BAC} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong m\u1ed9t tam gi\u00e1c) <br\/> $\\Rightarrow$ $\\widehat{BAC} = 180^{o} - (\\widehat{ABC} + \\widehat{ACB}) = 180^{o} - (75^{o} + 60^{o}) = 45^{o}$ (*) <br\/> T\u1eeb gi\u1ea3 thi\u1ebft ta c\u00f3: <br\/> $\\widehat{B_{2}} = \\widehat{ABC} - \\widehat{B_{1}} = 75^{o} - 45^{o} = 30^{o}$ <br\/> $\\widehat{C_{2}} = \\widehat{ACB} - \\widehat{C_{1}} = 60^{o} - 45^{o} = 15^{o}$ <br\/> $\\blacktriangleright$ Gi\u1ea3 s\u1eed $MA \\neq MB$ ta x\u00e9t hai tr\u01b0\u1eddng h\u1ee3p <br\/> $\\blacktriangleright$ $ MA < MB$ <br\/> X\u00e9t $\\triangle{MAB}$, v\u00ec $MA < MB$ n\u00ean $\\widehat{B_{2}} < \\widehat{A_{2}}$ (quan h\u1ec7 g\u00f3c - c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong $\\triangle{AMB}$) <br\/> V\u00ec $MC = MB$ ($\\triangle{BMC}$ c\u00e2n t\u1ea1i $M$) $\\Rightarrow MA < MC$ <br\/> Do \u0111\u00f3: $\\widehat{C_{2}} < \\widehat{A_{1}}$ (quan h\u1ec7 g\u00f3c - c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong $\\triangle{MAC}$) <br\/> $\\Rightarrow$ $\\widehat{B_{2}} + \\widehat{C_{2}} < \\widehat{A_{1}} + \\widehat{A_{2}}$ hay $30^{o} + 15^{o} < 45^{o} < \\widehat{BAC}$ tr\u00e1i v\u1edbi (*) <br\/> V\u1eady \u0111i\u1ec1u gi\u1ea3 s\u1eed l\u00e0 sai <br\/> $\\blacktriangleright$ $MA > MB$ <br\/> X\u00e9t $\\triangle{MAB}$, v\u00ec $MA > MB$ n\u00ean $\\widehat{B_{2}} > \\widehat{A_{2}}$ (quan h\u1ec7 g\u00f3c - c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong $\\triangle{AMB}$) <br\/> V\u00ec $MC = MB$ n\u00ean $MA > MC$ <br\/> Do \u0111\u00f3: $\\widehat{C_{2}} > \\widehat{A_{1}}$ (quan h\u1ec7 g\u00f3c - c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong $\\triangle{MAC}$) <br\/> Suy ra: $\\widehat{B_{2}} + \\widehat{C_{2}} > \\widehat{A_{1}} + \\widehat{A_{2}}$ hay $30^{o} + 15^{o} = 45^{o} > \\widehat{BAC}$ tr\u00e1i v\u1edbi (*) <br\/> V\u1eady \u0111i\u1ec1u gi\u1ea3 s\u1eed $MA \\neq MB$ l\u00e0 sai, hay $MA = MB$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span> ","column":2}]}],"id_ques":2034},{"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":"Cho tam gi\u00e1c $ABC$, g\u1ecdi $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AC$. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $IB$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $ID = IB$. G\u1ecdi $M$ v\u00e0 $N$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$ v\u00e0 $CD$. C\u00e1c \u0111\u01b0\u1eddng $AM, AN$ c\u1eaft $BD$ l\u1ea7n l\u01b0\u1ee3t \u1edf $G$ v\u00e0 $K$. G\u1ecdi $P$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB$. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang nh\u1ea5t? ","select":["A. $C, G, P$ th\u1eb3ng h\u00e0ng ","B. $BG = GK = KD$","C. $BG > GK = KD$ ","D. A v\u00e0 B \u0111\u1ec1u \u0111\u00fang"],"hint":"","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai24/lv3/img\/H7C3B24_K5.png' \/><\/center><br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABC}$, c\u00f3 $M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$, $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AC$ <br\/> $\\Rightarrow$ $AM, BI$ l\u00e0 c\u00e1c \u0111\u01b0\u1eddng trung tuy\u1ebfn, ch\u00fang c\u1eaft nhau t\u1ea1i $G$ <br\/> $\\Rightarrow$ $G$ l\u00e0 tr\u1ecdng t\u00e2m $\\triangle{ABC}$ n\u00ean $G$ thu\u1ed9c trung tuy\u1ebfn $CP$ hay ba \u0111i\u1ec3m $C, G, P$ th\u1eb3ng h\u00e0ng <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n A \u0111\u00fang <\/b> <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABC}$, ta c\u00f3: $BG = \\dfrac{2}{3}BI$ v\u00e0 $GI = \\dfrac{1}{2}BG$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) (1) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ACD}$ c\u00f3 $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AC$, $N$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $CD$ <br\/> $\\Rightarrow$ $AN, DI$ l\u00e0 c\u00e1c \u0111\u01b0\u1eddng trung tuy\u1ebfn, ch\u00fang c\u1eaft nhau t\u1ea1i $K$ <br\/> $\\Rightarrow$ $K$ l\u00e0 tr\u1ecdng t\u00e2m $\\triangle{ACD}$ (t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng trung tuy\u1ebfn trong tam gi\u00e1c) <br\/> $\\Rightarrow$ $KD = \\dfrac{2}{3}ID$ v\u00e0 $KI = \\dfrac{1}{2}KD$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) (2) <br\/> T\u1eeb (1) v\u00e0 (2) v\u00e0 gi\u1ea3 thi\u1ebft $IB = ID$ $\\Rightarrow$ $BG = GK = KD$ <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n B \u0111\u00fang, C sai <\/b> <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t l\u00e0: D <\/span> ","column":2}]}],"id_ques":2035},{"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":"N\u1ebfu m\u1ed9t tam gi\u00e1c c\u00f3 hai \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong b\u1eb1ng nhau th\u00ec tam gi\u00e1c \u0111\u00f3 c\u00e2n. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>?","select":["A. \u0110\u00daNG ","B. SAI"],"hint":"Gi\u1ea3 s\u1eed tam gi\u00e1c kh\u00f4ng c\u00e2n","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai24/lv3/img\/H7C3B24_K6.png' \/><\/center><br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABC}$ c\u00f3 hai \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong b\u1eb1ng nhau $BM = CN$. Ta s\u1ebd ch\u1ee9ng minh $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$ <br\/> Gi\u1ea3 s\u1eed $\\widehat{B} > \\widehat{C}$ v\u00e0 $BM, CN$ l\u00e0 c\u00e1c \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $\\Rightarrow$ $\\widehat{B_{1}} > \\widehat{C_{1}}$ (1) <br\/> Qua $M$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi $AB$, qua $N$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi $BM$, hai \u0111\u01b0\u1eddng th\u1eb3ng c\u1eaft nhau t\u1ea1i $D$ <br\/> $\\blacktriangleright$ $\\triangle{BNM} = \\triangle{DMN}$ (g.c.g) <br\/> $\\begin{cases} \\widehat{DNM} = \\widehat{BMN} (\\text{so} \\hspace{0,2cm} \\text{le} \\hspace{0,2cm} \\text{trong}) \\\\ MN \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{BMN} = \\widehat{DNM} (\\text{so} \\hspace{0,2cm} \\text{le} \\hspace{0,2cm} \\text{trong}) \\end{cases}$ <br\/> $\\Rightarrow$ $BM = ND$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng); $\\widehat{B_{2}} = \\widehat{D_{1}}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> Theo gi\u1ea3 thi\u1ebft $BM = CN$ n\u00ean $ND = NC$ <br\/> $\\Rightarrow$ $\\triangle{NCD}$ c\u00e2n t\u1ea1i $N$ $\\Rightarrow$ $\\widehat{NCD} = \\widehat{NDC}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) (2) <br\/> $\\blacktriangleright$ V\u00ec $\\widehat{B_{2}} = \\widehat{D_{1}}$ v\u00e0 $\\widehat{B_{2}} > \\widehat{C_{2}}$ ($\\widehat{B} > \\widehat{C}$ v\u00e0 $\\widehat{B_{1}} > \\widehat{C_{1}}$) <br\/> $\\Rightarrow$ $\\widehat{D_{1}} > \\widehat{C_{2}}$ (3) <br\/> T\u1eeb (2) v\u00e0 (3) $\\Rightarrow$ $\\widehat{D_{2}} < \\widehat{C_{3}}$ <br\/> Do \u0111\u00f3 $MC < MD = BN$ <br\/> Hai tam gi\u00e1c $BNC$ v\u00e0 $BMC$ c\u00f3 $BC$ chung, $CN = BM$ v\u00e0 $BN > CM$ <br\/> $\\Rightarrow$ $\\widehat{C_{1}} > \\widehat{B_{1}}$ (quan h\u1ec7 g\u00f3c - c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong hai tam gi\u00e1c) <br\/> M\u00e2u thu\u1eabn v\u1edbi (1), ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 c\u0169ng d\u1eabn \u0111\u1ebfn m\u00e2u thu\u1eabn <br\/> V\u1eady $\\widehat{B} = \\widehat{C}$ suy ra $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span> ","column":2}]}],"id_ques":2036},{"time":24,"part":[{"title":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u","title_trans":"Cho tam gi\u00e1c nh\u1ecdn $ABC$ v\u00e0 \u0111\u01b0\u1eddng cao $AH$, \u0111\u01b0\u1eddng trung tuy\u1ebfn $BM$, \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $CN$ c\u1eaft nhau t\u1ea1o th\u00e0nh tam gi\u00e1c $RPQ$. H\u00e3y s\u1eafp x\u1ebfp c\u00e1c \u00fd ch\u1ee9ng mih $\\triangle{RPQ}$ kh\u00f4ng l\u00e0 tam gi\u00e1c \u0111\u1ec1u. ","temp":"sequence","correct":[[[5],[2],[1],[4],[3]]],"list":[{"point":10,"image":"https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai24/lv3/img\/H7C3B24_K7.png","left":[" L\u1ea1i c\u00f3 $\\widehat{BAC} = 60^{o}$ n\u00ean l\u00e0 tam gi\u00e1c \u0111\u1ec1u <br\/> V\u1eady ba \u0111i\u1ec3m $R, P, Q$ tr\u00f9ng nhau, tr\u00e1i v\u1edbi gi\u1ea3 thi\u1ebft. V\u1eady $\\triangle{RPQ}$ kh\u00f4ng th\u1ec3 l\u00e0 tam gi\u00e1c \u0111\u1ec1u "," Do \u0111\u00f3 $\\widehat{ABC} = 2\\widehat{RCH} = 60^{o}$ ($CN$ l\u00e0 tia ph\u00e2n gi\u00e1c)"," Gi\u1ea3 s\u1eed ng\u01b0\u1ee3c l\u1ea1i $\\triangle{RPQ}$ \u0111\u1ec1u. <br\/> Khi \u0111\u00f3 trong tam gi\u00e1c vu\u00f4ng $CRH$ c\u00f3 $\\widehat{HRC} = 60^{o}$ ($\\triangle{RPQ}$ \u0111\u1ec1u) $\\Rightarrow$ $\\widehat{RCH} = 30^{o}$ "," $\\triangle{APM}$ c\u00f3 $\\widehat{PAM} = 30^{o}$, $\\widehat{APM} = 60^{o}$ n\u00ean $\\widehat{AMP} = 90^{o}$ hay $BM \\perp AC$ <br\/> $\\triangle{ABC}$ c\u00f3 \u0111\u01b0\u1eddng trung tuy\u1ebfn $BM$ l\u00e0 \u0111\u01b0\u1eddng cao n\u00ean l\u00e0 tam gi\u00e1c c\u00e2n "," $\\triangle{AHC}$ vu\u00f4ng t\u1ea1i $H$ c\u00f3 $\\widehat{ACH} = 60^{o}$ $\\Rightarrow$ $\\widehat{HAC} = 30^{o}$ "],"top":100,"hint":"","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai24/lv3/img\/H7C3B24_K7.png' \/><\/center> Gi\u1ea3 s\u1eed ng\u01b0\u1ee3c l\u1ea1i $\\triangle{RPQ}$ \u0111\u1ec1u. <br\/> Khi \u0111\u00f3 trong tam gi\u00e1c vu\u00f4ng $CRH$ c\u00f3 $\\widehat{HRC} = 60^{o}$ ($\\triangle{RPQ}$ \u0111\u1ec1u) $\\Rightarrow$ $\\widehat{RCH} = 30^{o}$ <br\/> Do \u0111\u00f3 $\\widehat{ABC} = 2\\widehat{RCH} = 60^{o}$ ($CN$ l\u00e0 tia ph\u00e2n gi\u00e1c) <br\/> $\\triangle{AHC}$ vu\u00f4ng t\u1ea1i $H$ c\u00f3 $\\widehat{ACH} = 60^{o}$ $\\Rightarrow$ $\\widehat{HAC} = 30^{o}$ <br\/> $\\triangle{APM}$ c\u00f3 $\\widehat{PAM} = 30^{o}$, $\\widehat{APM} = 60^{o}$ n\u00ean $\\widehat{AMP} = 90^{o}$ hay $BM \\perp AC$ <br\/> $\\triangle{ABC}$ c\u00f3 \u0111\u01b0\u1eddng trung tuy\u1ebfn $BM$ l\u00e0 \u0111\u01b0\u1eddng cao n\u00ean l\u00e0 tam gi\u00e1c c\u00e2n <br\/> L\u1ea1i c\u00f3 $\\widehat{BAC} = 60^{o}$ n\u00ean l\u00e0 tam gi\u00e1c \u0111\u1ec1u <br\/> V\u1eady ba \u0111i\u1ec3m $R, P, Q$ tr\u00f9ng nhau, tr\u00e1i v\u1edbi gi\u1ea3 thi\u1ebft. V\u1eady $\\triangle{RPQ}$ kh\u00f4ng th\u1ec3 l\u00e0 tam gi\u00e1c \u0111\u1ec1u <\/span> "}]}],"id_ques":2037},{"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 g\u00f3c nh\u1ecdn $xOy$ nh\u1ecdn, tr\u00ean $Ox$ l\u1ea5y \u0111i\u1ec3m $A$ v\u00e0 tr\u00ean $Oy$ l\u1ea5y \u0111i\u1ec3m $B$ sao cho $OA = OB$. \u0110\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $OB$ v\u00e0 $OA$ c\u1eaft nhau t\u1ea1i $I$ ","hint":"","column":2,"number_true":2,"select":["A. $OI$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{xOy}$ ","B. $\\widehat{HOI} = \\widehat{IAO} $","C. $OI$ l\u00e0 trung tr\u1ef1c c\u1ee7a $AB$ ","D. $\\widehat{OAB} > \\widehat{OBA}$ "],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai24/lv3/img\/H7C3B24_K8.png' \/><\/center> <br\/>$\\blacktriangleright$ X\u00e9t $\\triangle{AOI}$ v\u00e0 $\\triangle{OIB}$ c\u00f3: <br\/> $HI$ l\u00e0 trung tr\u1ef1c c\u1ee7a $OA$ (gt) n\u00ean $OI = IA$ (t\u00ednh ch\u1ea5t) <br\/> $KI$ l\u00e0 trung tr\u1ef1c c\u1ee7a $OB$ (gt) n\u00ean $OI = IB$ (t\u00ednh ch\u1ea5t) <br\/> Do \u0111\u00f3 $OI = IA = IB$ <br\/> V\u00ec $OA = OB$ (gt), suy ra $\\triangle{OIA} = \\triangle{OIB}$ (c.c.c) <br\/> $\\Rightarrow$ $\\widehat{O_{1}} = \\widehat{O_{2}}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> Do \u0111\u00f3 $OI$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{xOy}$ <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n A \u0111\u00fang <\/b> <br\/>$\\blacktriangleright$ $\\triangle{OIA}$ c\u00f3 $OI = IA$ n\u00ean c\u00e2n t\u1ea1i $I$ <br\/> $\\Rightarrow$ $\\widehat{HOI} = \\widehat{IAO}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n B \u0111\u00fang <\/b> <br\/> $\\blacktriangleright$ $\\triangle{AOB}$ c\u00f3 $HI, KI$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c <br\/> $\\Rightarrow$ $OI$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c (t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a tam gi\u00e1c) <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n C \u0111\u00fang <\/b> <br\/> $\\blacktriangleright$ $\\triangle{AOB}$ c\u00f3 $OA = OB$ n\u00ean c\u00e2n t\u1ea1i $A$ <br\/> $\\Rightarrow$ $\\widehat{OAB} = \\widehat{OBA}$ <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":2038},{"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":"Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{A} = 90^{o}$. Tr\u00ean c\u1ea1nh $AC$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $\\widehat{ABC} = 3\\widehat{ABD}$. Tr\u00ean c\u1ea1nh $AB$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $\\widehat{ACB} = 3 \\widehat{ACE}$. G\u1ecdi $F$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $DB$ v\u00e0 $CE$, $I$ l\u00e0 giao \u0111i\u1ec3m c\u00e1c tia ph\u00e2n gi\u00e1c c\u1ee7a $\\triangle{BFC}$. Trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau kh\u1eb3ng \u0111\u1ecbnh n\u00e0o \u0111\u00fang? ","select":["A. $\\widehat{BFC} = 90^{o} $ ","B. $\\widehat{BFC} = 120^{o}$ ","C. $\\triangle{DEI}$ l\u00e0 tam gi\u00e1c \u0111\u1ec1u ","D. B v\u00e0 C \u0111\u00fang "],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $\\widehat{BFC}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $EI; ED; DI$ \u0111\u1ec3 bi\u1ebft $\\triangle{EDI}$ \u0111\u1ec1u l\u00e0 \u0111\u00fang hay sai <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai24/lv3/img\/H7C3B21_K01.png' \/><\/center> <br\/> Ta c\u00f3: <br\/> $\\blacktriangleright$ $\\widehat{B} + \\widehat{C} = 90^{o}$ (v\u00ec $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$) <br\/> V\u00ec $\\widehat{ABC} = 3\\widehat{ABD}$; $\\widehat{ACB} = 3\\widehat{ACE}$ (gt) <br\/> $\\Rightarrow$ $\\widehat{FBC} + \\widehat{FCB} = \\dfrac{2}{3}(\\widehat{B} + \\widehat{C}) = \\dfrac{2}{3}. 90^{o} = 60^{o}$ <br\/> Trong $\\triangle{FBC}$ c\u00f3: $\\widehat{FCB} + \\widehat{FBC} + \\widehat{BFC} = 180^{o}$ (\u0111\u1ecbnh l\u00fd t\u1ed5ng ba g\u00f3c) <br\/> $ \\begin{align} \\Rightarrow \\widehat{BFC} &= 180^{o} - (\\widehat{FBC} + \\widehat{BFC}) \\\\ &= 180^{o} - 60^{o} \\\\ &= 120^{o}\\end{align} $ <br\/> $\\blacktriangleright$ V\u00ec $\\widehat{BFC} = 120^{o}$ $\\Rightarrow$ $\\widehat{BFE} = \\widehat{CFD} = 180^{o} - 120^{o} = 60^{o}$ (c\u00f9ng k\u1ec1 b\u00f9 v\u1edbi $\\widehat{BFC}$) <br\/> V\u00ec $I$ l\u00e0 giao \u0111i\u1ec3m $3$ tia ph\u00e2n gi\u00e1c c\u1ee7a $\\triangle{BFC}$ <br\/> $\\Rightarrow$ $FI$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BFC}$ <br\/> Do \u0111\u00f3 $\\widehat{BFI} = \\widehat{CFI} = \\dfrac{\\widehat{BFC}}{2} = \\dfrac{120^{o}}{2} = 60^{o}$ <br\/> $\\triangle{BFE} = \\triangle{BFI}$ (g.c.g) v\u00ec: <br\/> $\\begin{cases} \\widehat{BFE} = \\widehat{BFI} = 60^{o} (cmt) \\\\ BF \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{EBF} = \\widehat{FBI} (\\widehat{ABD} = \\dfrac{1}{3}\\widehat{ABC}; \\widehat{DBI} = \\dfrac{1}{2}\\widehat{DBC}) \\end{cases}$ <br\/> $\\Rightarrow$ $BI = BE$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\blacktriangleright$ $\\triangle{BED} = \\triangle{BID}$ (c.g.c) v\u00ec: <br\/> $\\begin{cases} BE = BI (cmt) \\\\ \\widehat{EBD} = \\widehat{IBD} (\\triangle{EBF} = \\triangle{IBF}) \\\\ BD \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $ED = DI$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> $\\blacktriangleright$ $\\triangle{CFD} = \\triangle{CFI}$ (g.c.g) v\u00ec: <br\/> $\\begin{cases} \\widehat{CFD} = \\widehat{CFI} = 60^{o} (cmt) \\\\ FC \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{ICF} = \\widehat{DCF} (\\widehat{ACB} = 3\\widehat{ACE}; \\widehat{ECI} = \\widehat{BCI}) \\end{cases}$ <br\/> $\\Rightarrow$ $CI = CD$ <br\/> $\\triangle{CDE} = \\triangle{CIE}$ (c.g.c) v\u00ec: <br\/> $\\begin{cases} CD = CI (cmt) \\\\ \\widehat{ICF} = \\widehat{DCF} (\\triangle{ICF} = \\triangle{DCF}) \\\\ FC \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $DE = EI$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (2) <br\/> T\u1eeb (1) v\u00e0 (2) suy ra $DE = EI = ID$ <br\/> $\\Rightarrow$ $\\triangle{EID}$ l\u00e0 tam gi\u00e1c \u0111\u1ec1u <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span>","column":2}]}],"id_ques":2039},{"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 tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$, c\u00e1c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a c\u00e1c c\u1ea1nh $AB$ v\u00e0 $AC$ c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng $BC$ t\u1ea1i $N$ v\u00e0 $M$ ($N$ v\u00e0 $M$ n\u1eb1m ngo\u00e0i \u0111o\u1ea1n th\u1eb3ng $BC$). Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $AM$ l\u1ea5y \u0111i\u1ec3m $P$ sao cho $AP = MB$, khi \u0111\u00f3 $AM = AN = PC$. <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\/> T\u01b0\u01a1ng quan h\u00ecnh v\u1ebd ta c\u00f3 th\u1ec3 th\u1ea5y $AM = AN = PC$, t\u1eeb \u0111\u00f3 ta c\u1ea7n \u0111i ch\u1ee9ng minh <br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $\\triangle{MAK} = \\triangle{NAI}$ t\u1eeb \u0111\u00f3 suy ra $AM = AN$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $\\triangle{ABM} = \\triangle{CAP}$ \u0111\u1ec3 suy ra $AM = PC$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai24/lv3/img\/H7C3B22_K04.png' \/><\/center> <br\/> G\u1ecdi $I; K$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB$ v\u00e0 $AC$, n\u1ed1i $A$ v\u1edbi $N$, $P$ v\u1edbi $C$ <br\/> $\\blacktriangleright$ $M$ n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AC$ n\u00ean $MA = MC$ (t\u00ednh ch\u1ea5t) $\\Rightarrow$ $\\triangle{AMC}$ c\u00e2n t\u1ea1i $M$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $N$ n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB$ n\u00ean $NA = NB$ (t\u00ednh ch\u1ea5t) $\\Rightarrow$ $\\triangle{ANB}$ c\u00e2n t\u1ea1i $N$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $\\blacktriangleright$ X\u00e9t hai tam gi\u00e1c vu\u00f4ng $AMK$ v\u00e0 $ANI$ c\u00f3: <br\/> $AI = \\dfrac{AB}{2} = \\dfrac{AC}{2} = AK$ (v\u00ec $AB = AC$)<br\/> $\\widehat{MAK} = \\widehat{C_{1}}$ ($\\triangle{AMC}$ c\u00e2n) <br\/> $\\widehat{BAN} = \\widehat{B_{1}}$ ($\\triangle{ANB}$ c\u00e2n) <br\/> M\u00e0 $\\widehat{B_{1}} = \\widehat{C_{1}}$ ($\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$) $\\Rightarrow$ $\\widehat{MAK} = \\widehat{BAN}$ hay $\\widehat{MAK} = \\widehat{IAN}$<br\/> $\\Rightarrow$ $\\triangle{MAK} = \\triangle{ANI}$ (c\u1ea1nh g\u00f3c vu\u00f4ng - g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow$ $AM = AN$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABM}$ v\u00e0 $\\triangle{CAP}$ c\u00f3: <br\/> $AB = AC$ (gt); $BM = AP$ (gt) <br\/> $\\widehat{MBA} + \\widehat{ABC} = 180^{o}$ (2) <br\/> $\\widehat{PAC} + \\widehat{MAC} = 180^{o}$ (3) <br\/> M\u00e0 $\\widehat{ABN} = \\widehat{MCA}$ ($\\triangle{ABC}$ c\u00e2n) <br\/> V\u00e0 $\\widehat{CAM} = \\widehat{MCA}$ ($\\triangle{AMC}$ c\u00e2n) <br\/> $\\Rightarrow$ $\\widehat{ABN} = \\widehat{MAC}$ (4) <br\/> T\u1eeb (2), (3), (4) $\\Rightarrow$ $\\widehat{MBA} = \\widehat{PAC}$ <br\/> $\\Rightarrow$ $\\triangle{ABM} = \\triangle{CAP}$ (c.g.c) $\\Rightarrow$ $AM = PC$ (5) <br\/> T\u1eeb (1) v\u00e0 (5) $\\Rightarrow$ $AM = AN = PC$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span> ","column":2}]}],"id_ques":2040}],"lesson":{"save":0,"level":3}}