{"segment":[{"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":[[["5"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Cho $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 $AB = 3cm$, $AC = 4cm$. G\u1ecdi $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AC$, $d$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a \u0111o\u1ea1n $AC$ v\u00e0 $M$ l\u00e0 \u0111i\u1ec3m t\u00f9y \u00fd tr\u00ean $d$. H\u00e3y t\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $MA + MB$ <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $Min_{(MA + MB)}$ = _input_ $cm$ ","hint":"Ch\u1ec9 ra $MA + MB$ l\u1edbn h\u01a1n ho\u1eb7c b\u1eb1ng m\u1ed9t gi\u00e1 tr\u1ecb kh\u00f4ng \u0111\u1ed5i, t\u1eeb \u0111\u00f3 t\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t","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 $MA = MC$, khi \u0111\u00f3 $MA + MB = MB + MC$ <br\/><b>B\u01b0\u1edbc 2:<\/b> So s\u00e1nh $MB + MC$ v\u1edbi $BC$ r\u1ed3i t\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $MA + MB$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai18/lv3/img\/H7C3B18_K01.png' \/><\/center><br\/> G\u1ecdi $J$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng $d$ v\u00e0 $BC$ <br\/> V\u00ec $AI = IC$ (gt) <br\/> $MI \\perp AC$ (gt) <br\/> $\\Rightarrow$ $MA = MC$ (H\u00ecnh chi\u1ebfu b\u1eb1ng nhau th\u00ec \u0111\u01b0\u1eddng xi\u00ean b\u1eb1ng nhau) <br\/> $\\Rightarrow$ $MA + MB = MB + MC$ (1) <br\/> M\u00e0 $MB + MC \\geq BC$ (b\u1ea5t \u0111\u1eb3ng th\u1ee9c h\u1ec7 \u0111i\u1ec3m) (2) <br\/> X\u00e9t $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$ c\u00f3: <br\/> $BC^{2} = AB^{2} + AC^{2} = 3^{2} + 4^{2} = 5^{2}$ (\u0111\u1ecbnh l\u00fd Pitago) <br\/> $\\Rightarrow$ $BC = 5(cm)$ (3) <br\/> T\u1eeb (1), (2) v\u00e0 (3) $\\Rightarrow$ $MA + MB \\geq 5$ <br\/> $\\Rightarrow$ $MA + MB$ nh\u1ecf nh\u1ea5t l\u00e0 b\u1eb1ng $5$ khi v\u00e0 ch\u1ec9 khi $MB + MC = BC$ <br\/> $\\Leftrightarrow$ $M$ n\u1eb1m tr\u00ean \u0111o\u1ea1n th\u1eb3ng $BC$ <br\/> $\\Leftrightarrow$ $M \\equiv J$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: 5 <\/span> <br\/> <span class='basic_green'> <i> Ch\u00fa \u00fd: \u0110\u1ec3 gi\u1ea3i b\u00e0i to\u00e1n c\u1ef1c tr\u1ecb h\u00ecnh h\u1ecdc \"T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a t\u1ed5ng $MA + MB$ \", ta ph\u1ea3i ch\u1ee9ng t\u1ecf $MA + MB$ l\u1edbn h\u01a1n ho\u1eb7c b\u1eb1ng m\u1ed9t s\u1ed1 kh\u00f4ng \u0111\u1ed5i v\u00e0 ph\u1ea3i ch\u1ec9 r\u00f5 d\u1ea5u b\u1eb1ng c\u00f3 th\u1ec3 x\u1ea3y ra. \u0110i\u1ec1u ki\u1ec7n x\u1ea3y ra d\u1ea5u b\u1eb1ng l\u00e0 r\u1ea5t quan tr\u1ecdng. <br\/> Ch\u1eb3ng h\u1ea1n v\u1edbi b\u00e0i to\u00e1n tr\u00ean ta lu\u00f4n c\u00f3: $MA + MB \\geq AB = 3$. Tuy nhi\u00ean d\u1ea5u \"=\" kh\u00f4ng th\u1ec3 x\u1ea3y ra v\u00ec $M$ kh\u00f4ng th\u1ec3 n\u1eb1m tr\u00ean \u0111o\u1ea1n th\u1eb3ng $AB$. V\u00ec v\u1eady kh\u00f4ng c\u00f3 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $MA + MB$ l\u00e0 3 <\/i> <\/span> "}]}],"id_ques":1851},{"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 $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$. G\u1ecdi $H$ l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a $A$ tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng $BC$. H\u00e3y so s\u00e1nh $AH + BC$ v\u1edbi $AB + AC$ <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $AH + BC$ _input_ $AB + AC$ ","hint":"Tr\u00ean $AC$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $CA = CE$, tr\u00ean $AB$ l\u1ea5y \u0111i\u1ec3m $F$ sao cho $AH = AF$","explain":"<span class='basic_left'><span class='basic_green'>Ph\u00e2n t\u00edch b\u00e0i to\u00e1n:<\/span><br\/> \u0110\u1ec3 so s\u00e1nh \u0111\u01b0\u1ee3c $AH + BC$ v\u1edbi $AB + AC$ ta c\u1ea7n v\u1ebd th\u00eam \u0111\u01b0\u1eddng ph\u1ee5 l\u00e0m xu\u1ea5t hi\u1ec7n \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c, c\u00e1c \u0111\u01b0\u1eddng xi\u00ean v\u00e0 h\u00ecnh chi\u1ebfu \u0111\u1ec3 \u00e1p d\u1ee5ng \u0111\u01b0\u1ee3c \u0111\u1ecbnh l\u00fd quan h\u1ec7 gi\u1eefa \u0111\u01b0\u1eddng xi\u00ean v\u00e0 \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c, \u0111\u01b0\u1eddng xi\u00ean v\u00e0 h\u00ecnh chi\u1ebfu. C\u1ee5 th\u1ec3 ta c\u1ea7n l\u1ea5y \u0111i\u1ec3m $E$ tr\u00ean $BC$ sao cho $CA = CE$, tr\u00ean $AB$ l\u1ea5y \u0111i\u1ec3m $F$ sao cho $AH = AF$. Khi \u0111\u00f3 b\u00e0i to\u00e1n chuy\u1ec3n v\u1ec1 so s\u00e1nh $BE$ v\u00e0 $BF$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai18/lv3/img\/H7C3B18_K02.png' \/><\/center><br\/> $\\blacktriangleright$ K\u1ebb $AH \\perp BC$, ($H \\in BC$) <br\/> Tr\u00ean tia $CB$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $CE = CA$ (1) <br\/> Tr\u00ean tia $AB$ l\u1ea5y \u0111i\u1ec3m l\u1ea5y \u0111i\u1ec3m $F$ sao cho $AF = AH$ <br\/> $\\blacktriangleright$ V\u00ec $AH \\perp BC$ n\u00ean \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c $CH$ nh\u1ecf h\u01a1n \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng xi\u00ean $CA$ <br\/> $\\Rightarrow$ $CH < CE$ (2) <br\/> M\u1eb7t kh\u00e1c do $CA \\perp AB$ n\u00ean $CA < CB$ (\u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c nh\u1ecf h\u01a1n \u0111\u01b0\u1eddng xi\u00ean) (3) <br\/> T\u1eeb (1), (2) v\u00e0 (3) $\\Rightarrow$ $CH < CE < CB$, suy ra $E$ n\u1eb1m gi\u1eefa $H$ v\u00e0 $B$ <br\/> V\u00ec $AH < AB$ (\u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c nh\u1ecf h\u01a1n \u0111\u01b0\u1eddng xi\u00ean) <br\/> $\\Rightarrow$ $AF < AB$ (v\u00ec $AH = AF$) <br\/> $\\Rightarrow$ \u0110i\u1ec3m $F$ n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m $A$ v\u00e0 $B$ <br\/> $\\blacktriangleright$ L\u1ea1i c\u00f3 $CE = CA$ $\\Rightarrow$ $\\triangle{ACE}$ c\u00e2n t\u1ea1i $C$ $\\Rightarrow$ $\\widehat{CAE} = \\widehat{E_{1}}$ <br\/> M\u00e0 $\\begin{cases} \\widehat{CAE} + \\widehat{A_{1}} = 90^{o} \\\\ \\widehat{E_{1}} + \\widehat{A_{2}} = 90^{o} \\end{cases}$ <br\/> $\\Rightarrow$ $\\widehat{A_{1}} = \\widehat{A_{2}}$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{AHE}$ v\u00e0 $\\triangle{AFE}$ c\u00f3: <br\/> $\\begin{cases} AE \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{A_{1}} = \\widehat{A_{2}} (cmt) \\\\ AH = AF (\\text{c\u00e1ch} \\hspace{0,2cm} \\text{l\u1ea5y} \\hspace{0,2cm} \\text{\u0111i\u1ec3m} F) \\end{cases}$ <br\/> $\\Rightarrow \\triangle{AHE} = \\triangle{AFE}$ (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: $AH + BC = AF + CE + BE$ (4) <br\/> $AB + AC = AF + BF + CE$ (5) <br\/> M\u00e0 $BE > BF$ (\u0111\u01b0\u1eddng xi\u00ean l\u1edbn h\u01a1n \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c) (6) <br\/> T\u1eeb (4), (5), (6) $\\Rightarrow$ $AH + BC > AB + AC$ <br\/> <span class='basic_pink'> V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0 d\u1ea5u: > <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: \u1ede b\u00e0i to\u00e1n n\u00e0y ta \u0111\u00e3 s\u1eed d\u1ee5ng m\u1ed9t ph\u01b0\u01a1ng ph\u00e1p quen thu\u1ed9c c\u1ee7a l\u1edbp $6$: \u0110\u1ec3 ch\u1ee9ng minh \u0111i\u1ec3m $E$ n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m $H$ v\u00e0 $B$ ta ch\u1ee9ng minh $CH < CE < CB$, t\u01b0\u01a1ng t\u1ef1 v\u1edbi \u0111i\u1ec3m $F$ <\/i> <\/span> "}]}],"id_ques":1852},{"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 $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$. T\u1eeb m\u1ed9t \u0111i\u1ec3m $E$ b\u1ea5t k\u00ec tr\u00ean c\u1ea1nh $AB$, k\u1ebb m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi \u0111\u00e1y $BC$, \u0111\u01b0\u1eddng th\u1eb3ng n\u00e0y c\u1eaft $AC$ t\u1ea1i $F$. Khi \u0111\u00f3 $BF = \\dfrac{EF + BC}{2}$. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":[" A. \u0110\u00daNG"," B. SAI"],"hint":"K\u1ebb $AH \\perp BC$ l\u00e0m xu\u1ea5t hi\u1ec7n \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c, \u0111\u01b0\u1eddng xi\u00ean v\u00e0 h\u00ecnh chi\u1ebfu \u0111\u1ec3 so s\u00e1nh","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> K\u1ebb $AH \\perp BC$, g\u1ecdi $J$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AH$ v\u00e0 $EF$, g\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BF$ v\u00e0 $AH$ <br\/> So s\u00e1nh $HB$ v\u00e0 $HC$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $JE$ v\u00e0 $JF$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> So s\u00e1nh $IB$ v\u1edbi $HB$, $IF$ v\u1edbi $JF$ r\u1ed3i so s\u00e1nh $BC$ v\u1edbi $\\dfrac{EF + BC}{2}$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai18/lv3/img\/H7C3B18_K03.png' \/><\/center> <br\/> $\\blacktriangleright$ K\u1ebb $AH \\perp BC$, do $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$ n\u00ean $\\widehat{B}$ v\u00e0 $\\widehat{C}$ nh\u1ecdn, suy ra \u0111i\u1ec3m $H$ n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m $B$ v\u00e0 $C$ <br\/> V\u00ec $EF \/\/ BC$ (gt) $\\Rightarrow$ $AH \\perp EF$, gi\u1ea3 s\u1eed t\u1ea1i \u0111i\u1ec3m $J$ <br\/> G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BF$ v\u00e0 $AH$ <br\/> $\\blacktriangleright$ Ta c\u00f3: $AB = AC$ ($\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$) <br\/> $\\Rightarrow$ $HB =HC$ (\u0111\u01b0\u1eddng xi\u00ean b\u1eb1ng nhau th\u00ec h\u00ecnh chi\u1ebfu b\u1eb1ng nhau) (1) <br\/> $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$ n\u00ean $\\widehat{B} = \\widehat{C}$ <br\/> $\\Rightarrow$ $\\widehat{AEF} = \\widehat{B} = \\widehat{C} = \\widehat{AFE}$ (\u0111\u1ed3ng v\u1ecb) <br\/> $\\Rightarrow$ $\\triangle{AEF}$ c\u00e2n t\u1ea1i $A$ $\\Rightarrow$ $AE = AF$ <br\/> $\\Rightarrow$ $JE = JF$ (\u0111\u01b0\u1eddng xi\u00ean b\u1eb1ng nhau th\u00ec h\u00ecnh chi\u1ebfu b\u1eb1ng nhau) (2) <br\/> $\\blacktriangleright$ M\u1eb7t kh\u00e1c ta c\u00f3: $BI > BH$ (\u0111\u01b0\u1eddng xi\u00ean l\u1edbn h\u01a1n \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c) (3) <br\/> $FI > FJ$ (\u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c l\u1edbn h\u01a1n h\u00ecnh chi\u1ebfu) (4) <br\/> T\u1eeb (1), (2), (3), (4) $\\Rightarrow$ $BI + IF > BH + JF = \\dfrac{1}{2}BC + \\dfrac{1}{2}EF$ <br\/> Hay $BF > \\dfrac{BC + EF}{2}$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh $BF = \\dfrac{EF + BC}{2}$ l\u00e0 <span class='basic_pink'>SAI <\/span> ","column":2}]}],"id_ques":1853},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u (\">\", \"<\" ho\u1eb7c \"=\") th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["="]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Cho hai \u0111o\u1ea1n th\u1eb3ng $AB$ v\u00e0 $CD$ thu\u1ed9c c\u00f9ng m\u1ed9t n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd c\u00f3 ch\u1ee9a \u0111\u01b0\u1eddng th\u1eb3ng $a$ v\u00e0 n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng th\u1eb3ng $a$ sao cho $AB = CD$ v\u00e0 $AB \/\/ CD$. H\u00ecnh chi\u1ebfu c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $AB$ v\u00e0 $CD$ tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng $a$ l\u00e0 $A'B'$ v\u00e0 $C'D'$. H\u00e3y so s\u00e1nh $A'B'$ v\u00e0 $C'D'$. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $A'B'$ _input_ $C'D'$ ","hint":"K\u1ebb $A'M \/\/ AB$, $C'N \/\/ CD$. X\u00e9t $\\triangle{MA'B'}$ v\u00e0 $\\triangle{NC'D'}$ \u0111\u1ec3 so s\u00e1nh $A'B'$ v\u00e0 $C'D'$","explain":"<span class='basic_left'><span class='basic_green'> H\u01b0\u1edbng d\u1eabn: <\/span> <br\/> <b> B\u01b0\u1edbc 1: <\/b> K\u1ebb $A'M \/\/ AB$, $C'N \/\/ CD$, sau \u0111\u00f3 ch\u1ee9ng minh $A'M = AB$, $C'N = CD$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $A'M = C'N$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> X\u00e9t $\\triangle{A'B'M}$ v\u00e0 $\\triangle{C'D'N}$, n\u1ebfu hai tam gi\u00e1c b\u1eb1ng nhau th\u00ec $A'B' = C'D'$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai18/lv3/img\/H7C3B18_K04.png' \/><\/center><br\/> $\\blacktriangleright$ Qua $A'$ k\u1ebb $A'M \/\/ AB$, qua $C'$ k\u1ebb $C'N \/\/ CD$ <br\/> N\u1ed1i $AM$ v\u00e0 $CN$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABM}$ v\u00e0 $\\triangle{MA'A}$ c\u00f3: <br\/> $\\begin{cases} \\widehat{A_{1}} = \\widehat{M_{2}} (\\text{so} \\hspace{0,2cm} \\text{le} \\hspace{0,2cm} \\text{trong}) \\\\ AM \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{A_{2}} = \\widehat{M_{1}} (\\text{so} \\hspace{0,2cm} \\text{le} \\hspace{0,2cm} \\text{trong}) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{ABM} = \\triangle{MA'A}$ (g.c.g) <br\/> $\\Rightarrow$ $AB = MA'$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{CDN}$ v\u00e0 $\\triangle{NC'C}$ c\u00f3: <br\/> $\\begin{cases} \\widehat{C_{1}} = \\widehat{N_{2}} (\\text{so} \\hspace{0,2cm} \\text{le} \\hspace{0,2cm} \\text{trong}) \\\\ CN \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{C_{2}} = \\widehat{N_{1}} (\\text{so} \\hspace{0,2cm} \\text{le} \\hspace{0,2cm} \\text{trong}) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{CDN} = \\triangle{NC'C}$ (g.c.g) <br\/> $\\Rightarrow$ $CD = C'N$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (2) <br\/> $\\blacktriangleright$ M\u00e0 $AB = CD$ (gt) (3) <br\/> T\u1eeb (1), (2), (3) $\\Rightarrow$ $MA' = C'N$ (4) <br\/> $\\blacktriangleright$ Ta c\u00f3: <br\/> $A'M \/\/ AB$ (c\u00e1ch v\u1ebd) <br\/> $C'N \/\/ CD$ (c\u00e1ch v\u1ebd) <br\/> M\u00e0 $AB \/\/ CD$ (gt) <br\/> Suy ra, $A'M \/\/ C'N$ (c\u00f9ng song song v\u1edbi $AB$ v\u00e0 $CD$) <br\/> $\\Rightarrow \\widehat{MA'B'} = \\widehat{NC'D'}$ (\u0111\u1ed3ng v\u1ecb) <br\/> X\u00e9t hai tam gi\u00e1c vu\u00f4ng $\\triangle{A'MB'}$ v\u00e0 $\\triangle{C'ND'}$ c\u00f3: <br\/> $\\begin{cases} A'M = C'N (\\text{theo} (4)) \\\\ \\widehat{MA'B'} = \\widehat{NC'D'} (\\text{cmt}) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{A'MB'} = \\triangle{NC'D'}$ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow A'B' = C'D'$ <br\/> <span class='basic_pink'> V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0 d\u1ea5u: = <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: Sai l\u1ea7m d\u1ec5 m\u1eafc ph\u1ea3i l\u00e0 khi ch\u1ec9 ra \u0111\u01b0\u1ee3c $A'M = C'N$ ta l\u1ea7m t\u01b0\u1edfng r\u1eb1ng hai \u0111\u01b0\u1eddng xi\u00ean b\u1eb1ng nhau th\u00ec hai h\u00ecnh chi\u1ebfu l\u00e0 $A'B'$ v\u00e0 $C'D'$ b\u1eb1ng nhau. <br\/> L\u01b0u \u00fd r\u1eb1ng: Khi x\u00e9t quan h\u1ec7 \u0111\u01b0\u1eddng xi\u00ean v\u00e0 h\u00ecnh chi\u1ebfu ch\u1ec9 x\u00e9t \u0111\u01b0\u1ee3c khi ch\u00fang c\u00f9ng xu\u1ea5t ph\u00e1t t\u1eeb m\u1ed9t \u0111i\u1ec3m b\u00ean ngo\u00e0i \u0111\u01b0\u1eddng th\u1eb3ng. <\/i> <\/span> "}]}],"id_ques":1854},{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["4"],["6"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Cho \u0111o\u1ea1n th\u1eb3ng $MN = 12cm$, $PQ = 8cm$ c\u1eaft nhau t\u1ea1i $O$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a m\u1ed7i \u0111o\u1ea1n v\u00e0 g\u00f3c t\u1ea1o th\u00e0nh gi\u1eefa hai \u0111o\u1ea1n \u1ea5y l\u00e0 $60^{o}$ $(\\widehat{NOQ} = 60^{o})$. T\u00ednh \u0111\u1ed9 d\u00e0i h\u00ecnh chi\u1ebfu c\u1ee7a $MN$ tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng $PQ$ v\u00e0 \u0111\u1ed9 d\u00e0i h\u00ecnh chi\u1ebfu c\u1ee7a $PQ$ tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng $MN$. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0:<\/b> _input_ $cm$; _input_$cm$ ","hint":"T\u1eeb c\u00e1c \u0111\u1ea7u m\u00fat c\u1ee7a c\u00e1c \u0111o\u1ea1n th\u1eb3ng h\u1ea1 \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c xu\u1ed1ng \u0111\u01b0\u1eddng th\u1eb3ng kia \u0111\u1ec3 t\u00ecm h\u00ecnh chi\u1ebfu, sau \u0111\u00f3 t\u00ednh \u0111\u1ed9 d\u00e0i c\u00e1c h\u00ecnh chi\u1ebfu","explain":"<span class='basic_left'><span class='basic_green'> H\u01b0\u1edbng d\u1eabn: <\/span> <br\/><b> B\u01b0\u1edbc 1: <\/b> T\u00ecm h\u00ecnh chi\u1ebfu c\u1ee7a $MN$ v\u00e0 $PQ$ b\u1eb1ng c\u00e1ch <br\/> H\u1ea1 $MH \\perp PQ$, $NK \\perp PQ$ <br\/> H\u1ea1 $PE \\perp MN$, $QF \\perp MN$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $OH = OK$, $OE = OF$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh $OH$ v\u00e0 $OE$ r\u1ed3i t\u00ednh $KH$ v\u00e0 $EF$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai18/lv3/img\/H7C3B18_K05.png' \/><\/center><br\/> $\\blacktriangleright$ H\u1ea1 $MH \\perp PQ$, $NK \\perp PQ$, suy ra $KH$ l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a $MN$ tr\u00ean $PQ$ <br\/> H\u1ea1 $PE \\perp MN$, $QF \\perp MN$, suy ra $EF$ l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a $PQ$ tr\u00ean $MN$ <br\/> $\\blacktriangleright$ $\\triangle{HOM} = \\triangle{KON}$ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) v\u00ec: <br\/> $\\begin{cases} \\widehat{OKN} = \\widehat{OHM} = 90^{o} (\\text{v\u00ec} NK \\perp PQ, MH \\perp PQ) \\\\ \\widehat{KON} = \\widehat{HOM} (\\text{\u0111\u1ed1i} \\hspace{0,2cm} \\text{\u0111\u1ec9nh}) \\\\ OM = ON (gt) \\end{cases}$ <br\/> $\\Rightarrow$ $OH = OK$ (1) <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u00f3 $OE = OF$ (2) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{MOH}$ c\u00f3 $\\widehat{MHO} = 90^{o}$ (theo c\u00e1ch v\u1ebd), $\\widehat{MOH} = 60^{o}$ (gt) <br\/> $\\Rightarrow \\widehat{OMH} = 180^{o} - 90^{o} - 60^{o} = 30^{o}$ (\u0111\u1ecbnh l\u00fd t\u1ed5ng $3$ g\u00f3c trong m\u1ed9t tam gi\u00e1c) <br\/> $OH$ l\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n v\u1edbi $\\widehat{OMH} = 30^{o}$, n\u00ean $OH = \\dfrac{1}{2}OM$ (\u0111\u1ecbnh l\u00fd) <br\/> $\\Rightarrow$ $OH = \\dfrac{1}{2}. \\dfrac{12}{2} = 3(cm)$ (3) <br\/> T\u1eeb (1) v\u00e0 (3) $\\Rightarrow$ $OH = OK = 3cm$ $\\Rightarrow$ $KH = 3 . 2 = 6(cm)$ <br\/> $\\blacktriangleright$ T\u01b0\u01a1ng t\u1ef1, x\u00e9t $\\triangle{POE}$ c\u00f3 $\\widehat{OEP} = 90^{o}$, $\\widehat{EOP} = 60^{o}$ <br\/> $\\Rightarrow$ $\\widehat{OPE} = 180^{o} - 90^{o} - 60^{o} = 30^{o}$ (\u0111\u1ecbnh l\u00fd t\u1ed5ng $3$ g\u00f3c trong m\u1ed9t tam gi\u00e1c) <br\/> $\\Rightarrow$ $OE = \\dfrac{1}{2}OP = \\dfrac{1}{2} . \\dfrac{8}{2} = 2(cm)$ (4) <br\/> T\u1eeb (2) v\u00e0 (4) $\\Rightarrow$ $OE = OF = 2cm$ $\\Rightarrow$ $EF = 2 . 2 = 4(cm)$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $4$ v\u00e0 $6$ <\/span> <br\/> <span class='basic_green'> <i> Ch\u1ee9ng minh \u0111\u1ecbnh l\u00fd: Trong tam $ABC$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 $\\widehat{C} = 30^{o}$ th\u00ec $AB = \\dfrac{1}{2}BC$ <br\/> Ch\u1ee9ng minh: <br\/> X\u00e9t $\\triangle{ABC}$ c\u00f3 $\\widehat{A} = 90^{o}$ (gt), $\\widehat{C} = 30^{o}$ (gt) <br\/> $\\Rightarrow$ $\\widehat{B} = 60^{o}$ (\u0111\u1ecbnh l\u00fd t\u1ed5ng $3$ g\u00f3c trong m\u1ed9t tam gi\u00e1c) <br\/> G\u1ecdi $M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$ $\\Rightarrow$ $MB = MC = \\dfrac{BC}{2}$ <br\/> Trong tam gi\u00e1c vu\u00f4ng \u0111\u01b0\u1eddng trung tuy\u1ebfn \u1ee9ng v\u1edbi c\u1ea1nh huy\u1ec1n b\u1eb1ng n\u1eeda c\u1ea1nh huy\u1ec1n <br\/> $\\Rightarrow$ $AM = \\dfrac{BC}{2} = BM = MC$ <br\/> X\u00e9t $\\triangle{ABM}$ c\u00f3 $AM = BM$ (cmt) v\u00e0 $\\widehat{B} = 60^{o}$ <br\/> $\\Rightarrow$ $\\triangle{ABM}$ l\u00e0 tam gi\u00e1c \u0111\u1ec1u. <br\/> $\\Rightarrow AB = AM = BM = \\dfrac{BC}{2}$ (\u0111pcm) <\/i> <\/span> "}]}],"id_ques":1855},{"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 $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$ v\u00e0 c\u00f3 $\\widehat{B} > \\widehat{C}$. G\u1ecdi $H$ l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a \u0111i\u1ec3m $A$ tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng $BC$. Tr\u00ean tia $BH$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $HD = HB$ (\u0111i\u1ec3m $D$ kh\u00e1c \u0111i\u1ec3m $B$). G\u1ecdi $E$ l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a $D$ tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng $AC$ v\u00e0 $K$ l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a $C$ tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng $AD$ . <br\/> Khi \u0111\u00f3 $DE = DK$<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> Ch\u1ee9ng minh \u0111i\u1ec3m $D$ n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m $H$ v\u00e0 $C$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $\\widehat{B} = \\widehat{CDE}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> Ch\u1ee9ng minh $\\widehat{CDE} = \\widehat{CDK}$ r\u1ed3i ch\u1ee9ng minh $DE = DK$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai18/lv3/img\/H7C3B18_K06.png' \/><\/center> <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABC}$ c\u00f3 $\\widehat{B} > \\widehat{C}$ (gt) <br\/> $\\Rightarrow$ $AC > AB$ (\u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c l\u1edbn h\u01a1n) <br\/> $\\Rightarrow$ $HC > HB$ (\u0111\u01b0\u1eddng xi\u00ean l\u1edbn h\u01a1n th\u00ec h\u00ecnh chi\u1ebfu l\u1edbn h\u01a1n) <br\/> M\u1eb7t kh\u00e1c $HB = HD$ (gt) $\\Rightarrow$ $HC > HD$, k\u1ebft h\u1ee3p v\u1edbi gi\u1ea3 thi\u1ebft $D \\in BH$ <br\/> Do \u0111\u00f3 $D \\in HC$ hay \u0111i\u1ec3m $D$ n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m $H$ v\u00e0 $C$ <br\/> $\\blacktriangleright$ V\u00ec $BA \\perp AC$ (gt) v\u00e0 $DE \\perp AC$ (gt) $\\Rightarrow$ $BA \/\/ DE$ (t\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song) <br\/> $\\Rightarrow$ $\\widehat{B} = \\widehat{CDE}$ (\u0111\u1ed3ng v\u1ecb) (1) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{AHB}$ v\u00e0 $\\triangle{AHD}$ c\u00f3: <br\/> $\\begin{cases} HB = HD (gt) \\\\ \\widehat{AHB} = \\widehat{AHD} = 90^{o} (AH \\perp BC) \\\\ AH \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{AHB} = \\triangle{AHC}$ (c.g.c) <br\/> $\\Rightarrow$ $\\widehat{B} = \\widehat{ADH}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) (2) <br\/> M\u1eb7t kh\u00e1c $\\widehat{ADH} = \\widehat{CDK}$ (\u0111\u1ed1i \u0111\u1ec9nh) <br\/> T\u1eeb (1), (2) $\\Rightarrow$ $\\widehat{CDE} = \\widehat{CDK}$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{CDE}$ v\u00e0 $\\triangle{CDK}$ c\u00f3: <br\/> $\\begin{cases} \\widehat{CDE} = \\widehat{CDK} (cmt) \\\\ DC \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{DEC} = \\widehat{DKC} = 90^{o} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{CDE} = \\triangle{CKD}$ (g.c.g) <br\/> $\\Rightarrow$ $DE = DK$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh $DE = DK$ l\u00e0 <span class='basic_pink'>\u0110\u00daNG <\/span> ","column":2}]}],"id_ques":1856},{"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 \u0111\u01b0\u1eddng th\u1eb3ng $d$ v\u00e0 \u0111i\u1ec3m $A$ kh\u00f4ng thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng $d$. $H$ l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a $A$ tr\u00ean $d$. V\u1ebd hai \u0111\u01b0\u1eddng xi\u00ean $AM$ v\u00e0 $AN$ c\u00f9ng n\u1eb1m tr\u00ean m\u1ed9t n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $AH$ sao cho $\\widehat{AMH} = 60^{o}$, $\\widehat{ANH} = 30^{o}$. Bi\u1ebft $AM = 7cm$, t\u00ednh \u0111\u1ed9 d\u00e0i h\u00ecnh chi\u1ebfu c\u1ee7a $AM$ v\u00e0 $AN$ tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng $d$ ","select":["A. $HM = 3,5cm; HN = 10cm$","B. $HM = 3,5cm; HN = 10,5cm$","C. $HM = 3cm; HN = 10,5cm$","D. $HM = 3cm; HN = 10cm$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> D\u1ef1a v\u00e0o \u0111\u1ecbnh l\u00fd: Trong tam gi\u00e1c vu\u00f4ng c\u00f3 m\u1ed9t g\u00f3c b\u1eb1ng $30^{o}$ th\u00ec c\u1ea1nh \u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c \u1ea5y b\u1eb1ng n\u1eeda c\u1ea1nh huy\u1ec1n \u0111\u1ec3 t\u00ednh $HM$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $AM = MN$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh $HN = HM + MN$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai18/lv3/img\/H7C3B18_K07.png' \/><\/center> <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{AHM}$ c\u00f3 $\\widehat{AMH} = 60^{o}$ (gt); $\\widehat{AHM} = 90^{o}$ (gt) <br\/> $\\Rightarrow$ $\\widehat{HAM} = 180^{o} - 90^{o} - 60^{o} = 30^{o}$ (\u0111\u1ecbnh l\u00fd t\u1ed5ng $3$ g\u00f3c trong m\u1ed9t tam gi\u00e1c) <br\/> C\u1ea1nh $HM$ \u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c $\\widehat{MAH} = 30^{o}$ <br\/> N\u00ean $HM = \\dfrac{1}{2} AM = \\dfrac{1}{2} . 7 = 3,5(cm)$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{AMN}$ c\u00f3 $\\widehat{AMH} = \\widehat{MAN} + \\widehat{MNA}$ (g\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c $AMN$) <br\/> $\\Rightarrow$ $\\widehat{MAN} = \\widehat{AMH} - \\widehat{MNA} = 60^{o} - 30^{o} = 30^{o}$ <br\/> $\\Rightarrow$ $\\triangle{MAN}$ c\u00e2n t\u1ea1i $M$ (do $\\widehat{MNA} = \\widehat{MAN} = 30^{o}$) <br\/> $\\Rightarrow$ $MN = AM = 7cm$ (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow$ $HN = HM + MN = 3,5cm + 7cm = 10,5cm$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B <\/span> ","column":2}]}],"id_ques":1857},{"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 $ABC$ c\u00f3 $\\widehat{A} = 90^{o}$, $\\widehat{B} = 54^{o}$. Tr\u00ean c\u1ea1nh $AC$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $\\widehat{DBC} = 18^{o}$. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $BD = AC$","B. $BD < AC$","C. $BD > AC$"],"hint":"V\u1ebd tia ph\u00e2n gi\u00e1c $BE$ c\u1ee7a $\\widehat{ABD}$, k\u1ebb $EF \\perp BD$. Ch\u1ee9ng minh $EF + EB > FD + BF$ ","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai18/lv3/img\/H7C3B18_K08.png' \/><\/center> <br\/> $\\blacktriangleright$ Ta c\u00f3: $\\widehat{B} = 54^{o}$; $\\widehat{DBC} = 18^{o}$ $\\Rightarrow$ $\\widehat{ABD} = 54^{o} - 18^{o} = 36^{o}$ <br\/> $\\triangle{ABC}$ c\u00f3 $\\widehat{A} = 90^{o}$, $\\widehat{B} = 54^{o}$ <br\/> $\\Rightarrow$ $\\widehat{C} = 180^{o} - 90^{o} - 54^{o} = 36^{o}$ (\u0111\u1ecbnh l\u00fd t\u1ed5ng $3$ g\u00f3c trong m\u1ed9t tam gi\u00e1c) <br\/> $\\blacktriangleright$ V\u1ebd tia ph\u00e2n gi\u00e1c $BE$ c\u1ee7a $\\widehat{ABD}$ <br\/> $\\Rightarrow$ $\\widehat{ABE} = \\widehat{EBD} = 18^{o}$ <br\/> T\u1eeb $E$ k\u1ebb $EF \\perp BD$ <br\/> X\u00e9t hai tam gi\u00e1c vu\u00f4ng $AEB$ v\u00e0 $FEB$ c\u00f3: <br\/> $\\begin{cases} \\text{C\u1ea1nh} \\hspace{0,2cm} \\text{huy\u1ec1n} \\hspace{0,2cm} BE \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{ABE} = \\widehat{FBE} (\\text{c\u00e1ch} \\hspace{0,2cm} \\text{l\u1ea5y} \\hspace{0,2cm} \\text{\u0111i\u1ec3m} \\hspace{0,2cm} E) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{ABE} = \\triangle{FBE}$ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow$ $AE = EF$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> $\\blacktriangleright$ $\\triangle{BEC}$ c\u00e2n t\u1ea1i $E$ (v\u00ec c\u00f3 $\\widehat{EBC} = \\widehat{EBD} + \\widehat{DBC} = 36^{o} = \\widehat{C}$) <br\/> $\\Rightarrow$ $EB = EC$ (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c c\u00e2n) (2) <br\/> $\\blacktriangleright$ T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $AC = AE + EC = EF + EB$ (3) <br\/> $\\triangle{ABD}$ c\u00f3 $\\widehat{ADB} = 180^{o} - 90^{o} - 36^{o} = 54^{o}$ <br\/> T\u1ee9c $\\widehat{EDF} = 54^{o}$ <br\/> $\\triangle{EDF}$ c\u00f3 $\\widehat{EDF} = 54^{o} > \\widehat{DEF} = 90^{o} - 54^{o} = 36^{o}$ <br\/> N\u00ean $EF > FD$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong tam gi\u00e1c $DEF$) (4) <br\/> $\\blacktriangleright$ Ta l\u1ea1i c\u00f3: $EB > BF$ (\u0111\u01b0\u1eddng xi\u00ean l\u1edbn h\u01a1n \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c) (5) <br\/> T\u1eeb (4) v\u00e0 (5) $\\Rightarrow$ $EF + EB > FD + BF = BD$ (6) <br\/> T\u1eeb (3) v\u00e0 (6) $\\Rightarrow$ $AC > BD$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B <\/span> ","column":3}]}],"id_ques":1858},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 $AB < AC$. Hai \u0111\u01b0\u1eddng cao $BD$ v\u00e0 $CE$. H\u00e3y so s\u00e1nh $BD$ v\u00e0 $CE$. ","select":["A. $BD = CE$","B. $BD > CE$","C. $BD < CE$"],"hint":"\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd: 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 \u0111\u1ec3 x\u00e9t $\\triangle{NBC}$ v\u00e0 $\\triangle{MCB}$ ","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai18/lv3/img\/H7C3B18_K09.png' \/><\/center> <br\/> $\\blacktriangleright$ Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $DB$ l\u1ea5y \u0111i\u1ec3m $M$ sao cho $DM = DB$ $\\Rightarrow$ $BM = 2BD$ <br\/> X\u00e9t $\\triangle{DMC}$ v\u00e0 $\\triangle{DBC}$ c\u00f3: <br\/> $\\begin{cases} MD = DB \\\\ \\widehat{BDC} = \\widehat{MDC} = 90^{o} (DB \\perp AC) \\\\ CD \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{DMC} = \\triangle{DBC}$ (c.g.c) <br\/> $\\Rightarrow$ $CM = CB$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\widehat{MCD} = \\widehat{BCD}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) $\\Rightarrow$ $\\widehat{BCM} = 2\\widehat{BCA}$ (1) <br\/> $\\blacktriangleright$ T\u01b0\u01a1ng t\u1ef1, tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $EC$ l\u1ea5y \u0111i\u1ec3m $N$ sao cho $EN = EC$ $\\Rightarrow$ $CN = 2CE$ <br\/> X\u00e9t $\\triangle{NEB}$ v\u00e0 $\\triangle{CEB}$ c\u00f3: <br\/> $\\begin{cases} NE = EC \\\\ \\widehat{NEB} = \\widehat{CEB} = 90^{o} (BE \\perp NC) \\\\ EB \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{NEB} = \\triangle{CEB}$ (c.g.c) <br\/> $\\Rightarrow$ $BC = NB$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\widehat{NBE} = \\widehat{CBE}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow$ $\\widehat{NBC} = 2\\widehat{CBA}$ (2) <br\/> $\\blacktriangleright$ M\u1eb7t kh\u00e1c ta c\u00f3: $AB < AC$ (gt) $\\Rightarrow$ $\\widehat{CBA} > \\widehat{BCA}$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong tam gi\u00e1c $ABC$) (3) <br\/> T\u1eeb (1), (2), (3) $\\Rightarrow$ $\\widehat{NBC} > \\widehat{MBC}$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{NBC}$ v\u00e0 $\\triangle{MCB}$ c\u00f3: <br\/> $\\begin{cases} NB = MC (= BC) \\\\ \\widehat{NBC} > \\widehat{MCB} (cmt) \\\\ BC \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $NC > BM$ (\u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c l\u1edbn h\u01a1n) <br\/> Suy ra, $CE > BD$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: Trong m\u1ed9t tam gi\u00e1c, \u0111\u01b0\u1eddng cao \u1ee9ng v\u1edbi c\u1ea1nh b\u00e9 h\u01a1n th\u00ec l\u1edbn h\u01a1n \u0111\u01b0\u1eddng cao \u1ee9ng v\u1edbi c\u1ea1nh l\u1edbn h\u01a1n. <\/i> <\/span> ","column":3}]}],"id_ques":1859},{"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 $A$. G\u1ecdi $D$ l\u00e0 \u0111i\u1ec3m t\u00f9y \u00fd n\u1eb1m trong tam gi\u00e1c $ABC$ sao cho $\\widehat{ADB} > \\widehat{ADC}$. <br\/> Khi \u0111\u00f3 $DC > DB$. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG","B. SAI"],"hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p ch\u1ee9ng minh b\u1eb1ng ph\u1ea3n ch\u1ee9ng. Gi\u1ea3 s\u1eed $DC \\leq DB$","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai18/lv3/img\/H7C3B18_K10.png' \/><\/center> <br\/> Gi\u1ea3 s\u1eed \u0111i\u1ec1u ng\u01b0\u1ee3c l\u1ea1i x\u1ea3y ra $DC \\leq DB$ <br\/> $\\blacktriangleright$ N\u1ebfu $DC = DB$ th\u00ec $\\triangle{ADB} = \\triangle{ADC}$ (c.c.c) v\u00ec: <br\/> $\\begin{cases} AB = AC (\\triangle{ABC} \\hspace{0,2cm} \\text{c\u00e2n}) \\\\ AD \\hspace{0,2cm} \\text{chung} \\\\ DB = DC \\end{cases}$ <br\/> $\\Rightarrow$ $\\widehat{ADB} = \\widehat{ADC}$ (tr\u00e1i v\u1edbi gi\u1ea3 thi\u1ebft l\u00e0 $\\widehat{ADB} = \\widehat{ADC}$) <br\/>$\\blacktriangleright$ N\u1ebfu $DC < DB$ th\u00ec trong $\\triangle{BCD}$ c\u00f3: <br\/> $\\widehat{B_{2}} < \\widehat{C_{2}}$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong tam gi\u00e1c $BCD$) <br\/> M\u00e0 $\\widehat{B_{1}} + \\widehat{B_{2}} = \\widehat{C_{1}} + \\widehat{C_{2}}$ (v\u00ec $\\widehat{B} = \\widehat{C}$) n\u00ean $\\widehat{C_{1}} < \\widehat{B_{1}}$ (1) <br\/> $\\triangle{ABD}$ v\u00e0 $\\triangle{ACD}$ c\u00f3: <br\/> $\\begin{cases} AB = AC (\\triangle{ABC} \\hspace{0,2cm} \\text{c\u00e2n}) \\\\ AD \\hspace{0,2cm} \\text{chung} \\\\ DC < DB \\end{cases}$ <br\/> $\\Rightarrow$ $\\widehat{A_{2}} < \\widehat{A_{1}}$ (\u0111\u1ed1i di\u1ec7n v\u1edbi c\u1ea1nh nh\u1ecf h\u01a1n) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $\\widehat{ADB} < \\widehat{ADC}$ (tr\u00e1i v\u1edbi gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow$ Kh\u00f4ng th\u1ec3 x\u1ea3y ra $DC \\leq BD$. <br\/> N\u00ean $DC > DB$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh $DC > DB$ l\u00e0 <span class='basic_pink'>\u0110\u00daNG <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: 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> ","column":2}]}],"id_ques":1860}],"lesson":{"save":0,"level":3}}