Bài tập nâng cao bài Định lí Ta - lét trong tam giác. Định lí đảo và hệ quả của định lí Ta

Home Lớp 8 Toán lớp 8 Định lí Ta - lét trong tam giác. Định lí đảo và hệ quả của định lí Ta - lét Bài tập nâng cao

{"segment":[{"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 h\u00ecnh v\u1ebd nh\u01b0 sau: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai14/lv3/img\/H8C3B1_K101.png' \/><\/center> <br\/> T\u00ednh $BN, MN$.<\/span>","select":["A. $BN = 3,75$; $MN = 2,25$","B. $BN = 3,25$; $MN = 2,75$","C. $BN = 2,25$; $MN = 3,75$","D. $BN = 60$; $MN = 36$"],"hint":"Ch\u1ee9ng minh $MN \/\/ AC$. T\u1eeb \u0111\u00f3 \u00e1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t \u0111\u1ec3 t\u00ednh $BN$ v\u00e0 $MN$.","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai14/lv3/img\/H8C3B1_K101.png' \/><\/center> <br\/> Ta c\u00f3: <br\/> $\\left.\\begin{array}{l} MN \\perp AB \\text{(gi\u1ea3 thi\u1ebft)}\\\\ AC \\perp AB \\text{(gi\u1ea3 thi\u1ebft)} \\end{array} \\right\\}$ $\\Rightarrow MN \/\/ AC$ (t\u00ednh ch\u1ea5t t\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song) <br\/> X\u00e9t $\\triangle ABC$ c\u00f3: $MN \/\/ AC$ (ch\u1ee9ng minh tr\u00ean) <br\/> $\\Rightarrow \\dfrac{BM}{BA}=\\dfrac{BN}{BC} = \\dfrac{MN}{AC}$ (h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t) <br\/> $\\Rightarrow \\dfrac{3}{12}=\\dfrac{BN}{15} = \\dfrac{MN}{9}$ <br\/> $\\Rightarrow$ $\\begin{cases}BN = \\dfrac{3.15}{12} = 3,75 \\\\ MN = \\dfrac{3.9}{12} = 2,25\\end{cases}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A. $BN = 3,75$ v\u00e0 $MN = 2,25$<\/span><\/span> <br\/> <\/span> ","column":2}]}],"id_ques":1671},{"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":"T\u00ecm \u0111\u1ed9 d\u00e0i c\u1ee7a $x$ trong h\u00ecnh v\u1ebd sau: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai14/lv3/img\/H8C3B1_K102.png' \/><\/center><\/span>","select":["A. $x \\approx 16,8$","B. $x \\approx 3,5$","C. $x \\approx 3,4$","D. $x \\approx 15,8$"],"hint":"Ch\u1ee9ng minh $EF \/\/ BC$. T\u1eeb \u0111\u00f3 \u00e1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t \u0111\u1ec3 t\u00ecm $x$.","explain":"H\u01b0\u1edbng d\u1eabn:<\/span> <br\/> B\u01b0\u1edbc 1:<\/b> T\u00ednh t\u1ec9 s\u1ed1 $\\dfrac{EA}{EB}$ v\u00e0 $\\dfrac{FA}{FC}$ <br\/> B\u01b0\u1edbc 2:<\/b> T\u1eeb t\u1ec9 s\u1ed1 t\u00ecm \u0111\u01b0\u1ee3c \u1edf b\u01b0\u1edbc 1, \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Ta-l\u00e9t \u0111\u1ea3o ch\u1ee9ng minh \u0111\u01b0\u1ee3c $EF$ \/\/ $BC$ <br\/> B\u01b0\u1edbc 3:<\/b> \u00c1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t t\u00ecm $x$.<\/span> <br\/> B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai14/lv3/img\/H8C3B1_K102.png' \/><\/center> <br\/> Ta c\u00f3: <br\/> $\\left.\\begin{array}{l} \\dfrac{EA}{EB} = \\dfrac{2}{5} \\\\ \\dfrac{FA}{FC} = \\dfrac{3}{7,5} = \\dfrac{2}{5} \\end{array} \\right\\}$ $\\Rightarrow \\dfrac{EA}{EB} = \\dfrac{FA}{FC}$ <br\/> $\\Rightarrow EF \/\/ BC$ (\u0111\u1ecbnh l\u00ed Ta-l\u00e9t \u0111\u1ea3o) <br\/> X\u00e9t $\\triangle ABC$ c\u00f3: $EF \/\/ BC$ (ch\u1ee9ng minh tr\u00ean) <br\/> $\\Rightarrow \\dfrac{AE}{AB}=\\dfrac{EF}{BC} $ (h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t) <br\/> $\\Rightarrow \\dfrac{AE}{AE + EB}=\\dfrac{EF}{BC} $ <br\/> $\\Rightarrow$ $\\dfrac{2}{2 + 5}=\\dfrac{x}{12} $ <br\/> $\\Rightarrow$ $\\dfrac{2}{7}=\\dfrac{x}{12} $ <br\/> $\\Rightarrow$ $x = \\dfrac{2.12}{7} \\approx 3,4$ <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C. $x \\approx 3,4$<\/span>","column":2}]}],"id_ques":1672},{"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":"T\u00ecm \u0111\u1ed9 d\u00e0i c\u1ee7a $x$ trong h\u00ecnh v\u1ebd sau: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai14/lv3/img\/H8C3B1_K103.png' \/><\/center><\/span>","select":["A. $x = 2,25$; $y = 12$","B. $x = 12$; $y = 3$","C. $x = 10$; $y = 2,25$","D. $x = 12$; $y = 2,25$"],"hint":"Ch\u1ee9ng minh $MN \/\/ PQ$. T\u1eeb \u0111\u00f3 \u00e1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t \u0111\u1ec3 t\u00ecm $x$ v\u00e0 $y$.","explain":"H\u01b0\u1edbng d\u1eabn:<\/span> <br\/> B\u01b0\u1edbc 1:<\/b> Ch\u1ee9ng minh $MN \/\/ PQ$ <br\/> B\u01b0\u1edbc 2:<\/b> \u00c1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t trong $\\triangle HPQ$ \u0111\u01b0\u1ee3c t\u1ec9 l\u1ec7 th\u1ee9c $\\dfrac{MN}{PQ} = \\dfrac{MH}{HQ} = \\dfrac{NH}{HP}$ <br\/> B\u01b0\u1edbc 3:<\/b> T\u1eeb t\u1ec9 l\u1ec7 th\u1ee9c tr\u00ean t\u00ecm \u0111\u01b0\u1ee3c $x$ v\u00e0 $y$.<\/span> <br\/> B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai14/lv3/img\/H8C3B1_K104.png' \/><\/center> <br\/> Ta c\u00f3: <br\/> $\\widehat{NMQ} = \\widehat{MQP}$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow MN \/\/ PQ$ (c\u1eb7p g\u00f3c so le trong b\u1eb1ng nhau) <br\/> X\u00e9t $\\triangle HPQ$ c\u00f3: <br\/> $MN \/\/ PQ$ (ch\u1ee9ng minh tr\u00ean) <br\/> $\\Rightarrow \\dfrac{MN}{PQ} = \\dfrac{MH}{HQ} = \\dfrac{NH}{HP}$ (h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t) <br\/> Hay $\\dfrac{4}{16} = \\dfrac{3}{x} = \\dfrac{y}{9}$ <br\/> $\\Rightarrow$ $\\begin{cases}x = \\dfrac{3.16}{4} = 12 \\\\ y = \\dfrac{4.9}{16} = 2,25\\end{cases}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D. $x = 12$; $y = 2,25$<\/span><\/span> <br\/> <\/span> ","column":2}]}],"id_ques":1673},{"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 h\u00ecnh thang $ABCD$ ($AB \/\/ CD$). Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $BA$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $BE = CD$. G\u1ecdi giao \u0111i\u1ec3m c\u1ee7a $AC$ v\u1edbi $BD$ v\u00e0 $DE$ theo th\u1ee9 t\u1ef1 l\u00e0 $H$ v\u00e0 $K$. So s\u00e1nh $\\dfrac{AK}{KC} = \\dfrac{AC}{CH}$.<\/span>","select":["A. $\\dfrac{AK}{KC}$ = $\\dfrac{AC}{CH}$","B. $\\dfrac{AK}{KC}$ > $\\dfrac{AC}{CH}$","C. $\\dfrac{AK}{KC}$ < $\\dfrac{AC}{CH}$"],"hint":"Khai th\u00e1c $BE = CD$; $AB \/\/ CD$ v\u00e0 v\u1eadn d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t. ","explain":"H\u01b0\u1edbng d\u1eabn:<\/span> <br\/> B\u01b0\u1edbc 1:<\/b> \u0110\u1eb7t $AB = a; BE = CD = b$ <br\/> B\u01b0\u1edbc 2:<\/b> T\u00ednh t\u1ec9 s\u1ed1 $\\dfrac{AK}{KC}$ theo $a$ v\u00e0 $b$ <br\/> B\u01b0\u1edbc 3:<\/b> T\u00ednh t\u1ec9 s\u1ed1 $\\dfrac{AC}{CH}$ theo $a$ v\u00e0 $b$. <br\/> B\u01b0\u1edbc 4:<\/b> T\u1eeb b\u01b0\u1edbc 2 v\u00e0 b\u01b0\u1edbc 3 t\u00ecm ra k\u1ebft qu\u1ea3 c\u1ea7n t\u00ecm.<\/span> <br\/> B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai14/lv3/img\/H8C3B1_K105.png' \/><\/center> <br\/> \u0110\u1eb7t $AB = a, BE = CD = b$ <br\/> ta c\u00f3: $AB \/\/ CD$ (gi\u1ea3 thi\u1ebft), $E \\ in AB$ <br\/> $\\Rightarrow AE \/\/ CD$ <br\/> $\\Rightarrow \\dfrac{AK}{KC} = \\dfrac{AE}{CD} = \\dfrac{a + b}{b} $ (1)<\/b> <br\/> L\u1ea1i c\u00f3: $AB \/\/ CD$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow \\dfrac{AH}{CH} = \\dfrac{AB}{CD} = \\dfrac{a}{b} $ <br\/> $\\Rightarrow \\dfrac{AH + CH}{CH} = \\dfrac{a + b}{b} $ <br\/> $\\Rightarrow \\dfrac{AC}{CH} = \\dfrac{a + b}{b} $ (2)<\/b> <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\dfrac{AK}{KC}$ = $\\dfrac{AC}{CH}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A. $\\dfrac{AK}{KC}$ = $\\dfrac{AC}{CH}$<\/span><\/span> ","column":3}]}],"id_ques":1674},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["="]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{A} = 120^o$, $AD$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c. So s\u00e1nh: $\\dfrac{1}{AB} + \\dfrac{1}{AC}$ v\u00e0 $\\dfrac{1}{AD}$ <br\/> \u0110\u00e1p \u00e1n:<\/b> $\\dfrac{1}{AB} + \\dfrac{1}{AC}$ _input_ $\\dfrac{1}{AD}$<\/span>","hint":" K\u1ebb th\u00eam \u0111\u01b0\u1eddng th\u1eb3ng qua $D$ v\u00e0 song song v\u1edbi $AB$.","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai14/lv3/img\/H8C3B1_K106.png' \/><\/center> <br\/> K\u1ebb $DK \/\/ AB$, ta c\u00f3: <br\/> $\\widehat{D_1} = \\widehat{A_1}$ (c\u1eb7p g\u00f3c so le trong) (1)<\/b> <br\/> L\u1ea1i c\u00f3: $AD$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a $\\triangle ABC$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow \\widehat{A_1} = \\widehat{A_2} = \\dfrac{1}{2} \\widehat{A} = \\dfrac{120^o}{2} = 60^o$ (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c) (2)<\/b> <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\widehat{A_2} = \\widehat{D_1} = 60^o$ <br\/> $\\Rightarrow \\triangle ADK$ \u0111\u1ec1u (\u0111\u1ecbnh ngh\u0129a) <br\/> $\\Rightarrow AD = AK = DK$ (t\u00edch ch\u1ea5t tam gi\u00e1c \u0111\u1ec1u) <br\/> X\u00e9t $\\triangle ABC$ c\u00f3: $DK \/\/ AB$ (c\u00e1ch v\u1ebd) <br\/> $\\Rightarrow \\dfrac{DK}{AB} = \\dfrac{CK}{AC}$ (h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t) <br\/> $\\Rightarrow \\dfrac{AD}{AB} = \\dfrac{CK}{AC}$ (v\u00ec $AD = DE$) (3)<\/b> <br\/> M\u1eb7t kh\u00e1c $\\dfrac{AD}{AC} = \\dfrac{CK}{AC}$ (v\u00ec $AD = AE$) (4)<\/b>) <br\/> T\u1eeb (3) v\u00e0 (4) $\\Rightarrow \\dfrac{AD}{AB} + \\dfrac{AD}{AC}$ = $\\dfrac{CK}{AC} + \\dfrac{AK}{AC} = \\dfrac{AC}{AC} = 1$ <br\/> $\\Rightarrow AD. \\left( \\dfrac{1}{AB} + \\dfrac{1}{AC}\\right) = 1$ <br\/> $\\Rightarrow$ $\\dfrac{1}{AB} + \\dfrac{1}{AC}$ = $\\dfrac{1}{AD}$ <br\/> V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 \"=\"<\/span> <br\/> Nh\u1eadn x\u00e9t:<\/span> Nh\u1eefng b\u00e0i to\u00e1n ch\u1ee9ng minh \u0111\u1eb3ng th\u1ee9c c\u00f3 ngh\u1ecbch \u0111\u1ea3o \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng, ch\u00fang ta n\u00ean bi\u1ebfn \u0111\u1ed5i v\u00e0 ch\u1ee9ng minh h\u1ec7 th\u1ee9c t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u00f3 t\u1ec9 s\u1ed1 c\u1ee7a hai \u0111o\u1ea1n th\u1eb3ng.<\/span>"}]}],"id_ques":1675},{"time":24,"part":[{"title":"Ch\u1ecdn 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"]],"list":[{"point":10,"img":"","ques":"Cho $\\triangle ABC$. T\u1eeb \u0111i\u1ec3m $E$ thu\u1ed9c c\u1ea1nh $AB$, k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi c\u1ea1nh $BC$, \u0111\u01b0\u1eddng n\u00e0y c\u1eaft c\u1ea1nh $AC$ t\u1ea1i \u0111i\u1ec3m $F$. Bi\u1ebft $AE = 4cm$; $EB = 8cm$ v\u00e0 $EF = 4cm$; $FC = 6cm$. <br\/> H\u00e3y ch\u1ecdn c\u00e1c \u0111\u00e1p \u00e1n \u0111\u00fang trong c\u00e1c \u0111\u00e1p \u00e1n sau:<\/b><\/span>","hint":"","column":2,"number_true":2,"select":["$BC = 12cm$","$AF = 3cm$"," $AC = 10cm$","$BC = 8cm$"],"explain":"H\u01b0\u1edbng d\u1eabn:<\/span> <br\/> B\u01b0\u1edbc 1:<\/b> T\u00ednh $AB$ <br\/> B\u01b0\u1edbc 2:<\/b> \u00c1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t \u0111\u1ec3 t\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh $BC, AF$ <br\/> B\u01b0\u1edbc 3:<\/b>T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh $AC$ <\/span> <br\/> B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai14/lv3/img\/H8C3B1_K107.png' \/><\/center> <br\/> Ta c\u00f3: $AB = AE + EB = 4 + 8 = 12$ ($cm$) <br\/> X\u00e9t $\\triangle ABC$ c\u00f3: $EF \/\/ BC$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow \\dfrac{AE}{AB} = \\dfrac{AF}{AC} = \\dfrac{EF}{BC}$ (h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t) <br\/> $\\Rightarrow \\dfrac{4}{12} = \\dfrac{AF}{AF + 6} = \\dfrac{4}{BC}$ <br\/> $\\Rightarrow \\begin{cases}BC = \\dfrac{4.12}{4} = 12 \\text{(cm)} \\\\ \\dfrac{AF}{AF + 6} = \\dfrac{1}{3} \\end{cases}\\Rightarrow 3AF = AF + 6 \\Rightarrow 2AF = 6 \\Rightarrow AF = 3 \\text{(cm)}$ <br\/> L\u1ea1i c\u00f3: $AC = AF + FC = 3 + 6 = 9 \\text{(cm)}$ <br\/> V\u1eady c\u00e1c \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: $BC = 12cm$; $AF = 3cm$<\/span> "}]}],"id_ques":1676},{"time":24,"part":[{"title":"Ch\u1ecdn nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["1","3","4"]],"list":[{"point":10,"img":"","ques":"Cho h\u00ecnh thang $ABCD$ ($AB \/\/ CD$). Qua giao \u0111i\u1ec3m $H$ c\u1ee7a hai \u0111\u01b0\u1eddng ch\u00e9o, k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi hai \u0111\u00e1y. \u0110\u01b0\u1eddng th\u1eb3ng n\u00e0y c\u1eaft c\u1ea1nh b\u00ean $AD$ v\u00e0 $BC$ l\u1ea7n l\u01b0\u1ee3t t\u1ea1i $E$ v\u00e0 $F$. <br\/> H\u00e3y ch\u1ecdn c\u00e1c \u0111\u00e1p \u00e1n \u0111\u00fang trong c\u00e1c \u0111\u00e1p \u00e1n sau:<\/b><\/span>","hint":"","column":2,"number_true":2,"select":["$H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $EF$","$H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BD$"," $\\dfrac{HF}{DC} = \\dfrac{BH}{BD}$","$\\dfrac{HE}{DC} = \\dfrac{AH}{AC}$"],"explain":"H\u01b0\u1edbng d\u1eabn:<\/span> <br\/> B\u01b0\u1edbc 1:<\/b> \u00c1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t v\u00e0o $\\triangle ADC$ <br\/> B\u01b0\u1edbc 2:<\/b> \u00c1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t v\u00e0o $\\triangle BDC$ <br\/> B\u01b0\u1edbc 3:<\/b>\u00c1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t v\u00e0o $\\triangle AHB$ <\/span> <br\/> B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai14/lv3/img\/H8C3B1_K108.png' \/><\/center> <br\/> X\u00e9t $\\triangle ADC$ c\u00f3: $EH \/\/ DC$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow \\dfrac{HE}{DC} = \\dfrac{AH}{AC} $ (h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t) (1)<\/b> <br\/> X\u00e9t $\\triangle BDC$ c\u00f3: $FH \/\/ DC$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow \\dfrac{HF}{DC} = \\dfrac{BH}{BD} $ (h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t) (2)<\/b> <br\/> X\u00e9t $\\triangle AHB$ c\u00f3: $AB \/\/ DC$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow \\dfrac{BH}{BD} = \\dfrac{AH}{AC} $ (h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t) (3)<\/b> <br\/> T\u1eeb (1), (2) v\u00e0 (3) ta \u0111\u01b0\u1ee3c: $\\dfrac{HE}{DC} = \\dfrac{HF}{DC} $ <br\/> $\\Rightarrow$ $HE = HF$ hay $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $EF$<\/span>"}]}],"id_ques":1677},{"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 $\\triangle ABC$ c\u00f3 $\\widehat{A} = 120^o$, $AB = 4cm, AC = 6cm$. T\u00ednh \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $AD$<\/span>","select":["A. $8cm$","B. $10cm$","C. $12cm$","D. $14cm$"],"hint":"K\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng qua $D$ song song v\u1edbi $AC$.","explain":"H\u01b0\u1edbng d\u1eabn:<\/span> <br\/> B\u01b0\u1edbc 1:<\/b> K\u1ebb $DE \/\/ AC$ <br\/> B\u01b0\u1edbc 2:<\/b> Ch\u1ee9ng minh $\\triangle ADE$ \u0111\u1ec1u <br\/> B\u01b0\u1edbc 3:<\/b> \u00c1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t t\u00ednh \u0111\u1ed9 d\u00e0i $DE$ <br\/> B\u01b0\u1edbc 4:<\/b> T\u1eeb $DE$ suy ra \u0111\u1ed9 d\u00e0i c\u1ea1nh $AD$.<\/span> <br\/> B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai14/lv3/img\/H8C3B1_K109.png' \/><\/center> <br\/> K\u1ebb $DE \/\/ AB$ $\\Rightarrow \\widehat{D_1} = \\widehat{A_1}$ (c\u1eb7p g\u00f3c so le trong) (1)<\/b> <br\/> L\u1ea1i c\u00f3: $AD$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a $\\triangle ABC$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow \\widehat{A_1} = \\widehat{A_2} = \\dfrac{A}{2} = \\dfrac{120^o}{2} = 60^o$ (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c) (2)<\/b> <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\widehat{D_1} = \\widehat{A_2} = 60^o$ <br\/> $\\Rightarrow$ $\\triangle ADE$ \u0111\u1ec1u <br\/> $\\Rightarrow$ $AD = AE = DE$ (t\u00ednh ch\u1ea5t tam gi\u00e1c \u0111\u1ec1u) <br\/> X\u00e9t $\\triangle ABC$ c\u00f3: $DE \/\/ AB$ (c\u00e1ch v\u1ebd) <br\/> $\\Rightarrow \\dfrac{DE}{AB} = \\dfrac{CE}{CA} $ (h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t) <br\/> $\\Rightarrow \\dfrac{DE}{4} = \\dfrac{CA - AE}{CA}$ <br\/> $\\Rightarrow \\dfrac{DE}{4} = \\dfrac{CA - DE}{CA}$ (v\u00ec AE = DE) <br\/> $\\Rightarrow \\dfrac{DE}{4} = \\dfrac{6 - DE}{6}$<br\/> $\\Leftrightarrow 6DE = 4(6 - DE)$ <br\/> $\\Leftrightarrow 6DE = 24 - 4DE$<br\/> $\\Leftrightarrow 2DE = 24$ <br\/>$\\Leftrightarrow DE = 12 \\text{(cm)}$ <br\/> L\u1ea1i c\u00f3: $DE = AD$ (ch\u1ee9ng minh tr\u00ean) <br\/> $\\Rightarrow AD = 12 \\text{(cm)}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C. $12cm$<\/span><\/span> <br\/> <\/span> ","column":2}]}],"id_ques":1678},{"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 h\u00ecnh 1, bi\u1ebft $ \\dfrac{AD}{AB} = \\dfrac{AE}{AC} $, $M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$. Ch\u1ee9ng minh \u0111\u01b0\u1ee3c:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai14/lv3/img\/H8C3B1_K109.png' \/><\/center> ","select":["A. $DN \/\/ BC$","B. $N$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $DE$","C. $\\dfrac{AN}{MN} = \\dfrac{AD}{BD} $","D. T\u1ea5t c\u1ea3 c\u00e1c \u00fd tr\u00ean \u0111\u1ec1u \u0111\u00fang"],"hint":"","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai14/lv3/img\/H8C3B1_K109.png' \/><\/center> <br\/> X\u00e9t $\\triangle ABC$ c\u00f3: $ \\dfrac{AD}{AB} = \\dfrac{AE}{AC} $ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow DE \/\/ BC$ (\u0111\u1ecbnh l\u00ed Ta-l\u00e9t \u0111\u1ea3o) <br\/> M\u00e0 $N \\in BC$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow DN \/\/ BC$ (\u00fd A \u0111\u00fang)<\/b> <br\/> X\u00e9t $\\triangle ABM$ c\u00f3: DN \/\/ BM (v\u00ec $M \\in BC$) <br\/> $\\Rightarrow \\begin{cases}\\dfrac{AD}{AN} = \\dfrac{AN}{MN} \\\\ \\dfrac{DN}{BM} = \\dfrac{AN}{AM} (1)\\end{cases} $ (h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t) <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta \u0111\u01b0\u1ee3c: $\\dfrac{AN}{AM} = \\dfrac{NE}{MC}$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\dfrac{DN}{BM} = \\dfrac{NE}{MC}$ <br\/> L\u1ea1i c\u00f3: $BM = MC$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow DN = NE$ hay $N$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $DE$ <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t l\u00e0: D. T\u1ea5t c\u1ea3 c\u00e1c \u00fd tr\u00ean \u0111\u1ec1u \u0111\u00fang<\/span><\/span> ","column":1}]}],"id_ques":1679},{"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 bi\u1ebft \u0111\u1ed9 d\u00e0i c\u1ee7a $AB$ g\u1ea5p $9$ l\u1ea7n \u0111\u1ed9 d\u00e0i c\u1ea1nh $CD$ v\u00e0 \u0111\u1ed9 d\u00e0i c\u1ee7a $A\u2019B\u2019$ g\u1ea5p $15$ l\u1ea7n \u0111\u1ed9 d\u00e0i c\u1ee7a $CD$. T\u00ednh t\u1ec9 s\u1ed1 c\u1ee7a hai \u0111o\u1ea1n th\u1eb3ng $AB$ v\u00e0 $A\u2019B\u2019$. ","select":[" A. $\\dfrac{3}{4}$ "," B. $\\dfrac{3}{5}$","C. $\\dfrac{5}{3}$","D. $\\dfrac{5}{4}$"],"explain":" H\u01b0\u1edbng d\u1eabn:<\/span> <br\/> S\u1eed d\u1ee5ng \u0111\u1ecbnh ngh\u0129a t\u1ec9 s\u1ed1 c\u1ee7a hai \u0111o\u1ea1n th\u1eb3ng<\/span> <br\/> B\u00e0i gi\u1ea3i:<\/span> <br\/> Theo \u0111\u1ec1 b\u00e0i ta c\u00f3: <br\/> $AB = 9.CD$ <br\/> $A\u2019B\u2019 = 15.CD$ <br\/> T\u1ec9 s\u1ed1 c\u1ee7a hai \u0111o\u1ea1n th\u1eb3ng $AB$ v\u00e0 $A\u2019B\u2019$ l\u00e0: <br\/> $\\dfrac{AB}{A'B'}=\\dfrac{9.CD}{15.CD}=\\dfrac{3}{5}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B. <\/span>","column":2}]}],"id_ques":1680}],"lesson":{"save":0,"level":3}}

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Câu hỏi này theo dạng chọn đáp án đúng, sau khi đọc xong câu hỏi, bạn bấm vào một trong số các đáp án mà chương trình đưa ra bên dưới, sau đó bấm vào nút gửi để kiểm tra đáp án và sẵn sàng chuyển sang câu hỏi kế tiếp

Trả lời đúng trong khoảng thời gian quy định bạn sẽ được + số điểm như sau:
Trong khoảng 1 phút đầu tiên	+ 5 điểm
Trong khoảng 1 phút -> 2 phút	+ 4 điểm
Trong khoảng 2 phút -> 3 phút	+ 3 điểm
Trong khoảng 3 phút -> 4 phút	+ 2 điểm
Trong khoảng 4 phút -> 5 phút	+ 1 điểm

Quá 5 phút: không được cộng điểm

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