{"segment":[{"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":" <span class='basic_left'> Cho h\u00ecnh thoi $ABCD$, tr\u00ean c\u1ea1nh $AD$ l\u1ea5y \u0111i\u1ec3m $M$. \u0110\u01b0\u1eddng th\u1eb3ng $CM$ c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng $AB$ t\u1ea1i $N.$ So s\u00e1nh $AB^2$ v\u00e0 $DM.BN$.<br\/><\/span>","select":[" A. $AB^2 > DM.BN$"," B. $AB^2 < DM.BN$","C. $AB^2 = DM.BN$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai16/lv3/img\/H8C3B3_K101.png' \/><\/center><br\/>$ABCD$ l\u00e0 h\u00ecnh thoi (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow$ $\\begin{cases}AB \/\/ CD; BC \/\/ AD \\\\ AB = CD = BC = AD \\end{cases}$ (t\u00ednh ch\u1ea5t h\u00ecnh thoi)<br\/>L\u1ea1i c\u00f3: $M \\in AD; N \\in AB$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow$ $\\begin{cases}AD \/\/ BC \\Rightarrow \\widehat{NAM} = \\widehat{ABC} \\text{(c\u1eb7p g\u00f3c \u0111\u1ed3ng v\u1ecb)} \\\\ AN \/\/ CD \\Rightarrow \\widehat{ANM} = \\widehat{MCD} \\text{(c\u1eb7p g\u00f3c so le trong)} \\end{cases}$<br\/>X\u00e9t $\\triangle{NAM}$ v\u00e0 $\\triangle{NBC}$ c\u00f3: <br\/>$\\widehat{NAM} = \\widehat{ABC}$ (c\u1eb7p g\u00f3c \u0111\u1ed3ng v\u1ecb)<br\/>$\\widehat{N_1}$ chung<br\/>$\\Rightarrow$ $\\triangle{NAM} \\backsim \\triangle{NBC}$ (g\u00f3c-g\u00f3c)<br\/>$\\Rightarrow \\dfrac{NA}{NB} = \\dfrac{AM}{BC}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7)<br\/>$\\Rightarrow \\dfrac{NA}{AM} = \\dfrac{NB}{BC}$ hay $\\dfrac{NA}{AM} = \\dfrac{NB}{AB}$ (v\u00ec $AB = BC$) <b>(1)<\/b> <br\/>X\u00e9t $\\triangle{NAM}$ v\u00e0 $\\triangle{CDM}$ c\u00f3: <br\/>$\\widehat{ANM} = \\widehat{MCD}$ (ch\u1ee9ng minh tr\u00ean)<br\/>$\\widehat{M_1} = \\widehat{M_2}$ (\u0111\u1ed1i \u0111\u1ec9nh)<br\/>$\\Rightarrow$ $\\triangle{NAM} \\backsim \\triangle{CDM}$ (g\u00f3c-g\u00f3c)<br\/>$\\Rightarrow \\dfrac{NA}{CD} = \\dfrac{AM}{DM}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7)<br\/>$\\Rightarrow \\dfrac{NA}{AM} = \\dfrac{CD}{DM}$ hay $\\dfrac{NA}{AM} = \\dfrac{AB}{DM}$ (v\u00ec $AB = CD$) <b>(2)<\/b><br\/>T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\dfrac{NB}{AB} = \\dfrac{AB}{DM}$ $\\Rightarrow AB^2 = DM.BN$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>C. $AB^2 = DM.BN$<\/span><br\/><span class='basic_left'><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>\u0110\u1ec3 ch\u1ee9ng minh \u0111\u1eb3ng th\u1ee9c t\u00edch, th\u00f4ng th\u01b0\u1eddng ch\u00fang ta bi\u1ebfn \u0111\u1ed5i ch\u00fang d\u01b0\u1edbi d\u1ea1ng t\u1ec9 l\u1ec7 th\u1ee9c v\u00e0 ch\u1ee9ng minh t\u1ec9 l\u1ec7 th\u1ee9c \u1ea5y.<\/span> <br\/><br\/>","column":3}]}],"id_ques":1730},{"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","4"]],"list":[{"point":10,"img":"","ques":"Cho c\u00e1c c\u1eb7p tam gi\u00e1c c\u00f3 c\u00e1c c\u1ea1nh c\u00f3 \u0111\u1ed9 d\u00e0i sau \u0111\u00e2y<br\/><b> H\u00e3y ch\u1ecdn nh\u1eefng c\u1eb7p tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng.<\/b>","hint":"","column":2,"number_true":2,"select":["A. 1,5cm; 2cm; 3cm v\u00e0 4,5cm; 6cm; 9cm","B. 2cm; 6cm; 9cm v\u00e0 4cm; 13cm; 18cm ","C. 3cm; 4cm; 5cm v\u00e0 6cm; 8cm; 12cm","D. 1,5cm; 3cm; 4cm v\u00e0 3cm; 6cm; 8cm"],"explain":"\u0110\u00e1p \u00e1n $A$ \u0111\u00fang v\u00ec: $\\dfrac{1,5}{4,5} = \\dfrac{2}{6} = \\dfrac{3}{9}$ <br\/>\u0110\u00e1p \u00e1n $B$ sai v\u00ec: $\\dfrac{2}{4} = \\dfrac{9}{18} \\neq \\dfrac{6}{13}$<br\/>\u0110\u00e1p \u00e1n $C$ sai v\u00ec: $\\dfrac{3}{6} = \\dfrac{4}{8} \\neq \\dfrac{5}{12}$<br\/>\u0110\u00e1p \u00e1n $D$ \u0111\u00fang v\u00ec: $\\dfrac{1,5}{3} = \\dfrac{3}{6} = \\dfrac{4}{8}$"}]}],"id_ques":1731},{"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":[[["20"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'>Cho h\u00ecnh v\u1ebd nh\u01b0 sau: <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai16/lv3/img\/H8C3B4_D114.png' \/><\/center><br\/> T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh $AC$.<br\/><b>\u0110\u00e1p \u00e1n:<\/b> $AC = $ _input_ ($cm$)<\/span>","hint":"Ch\u1ee9ng minh $\\triangle$ vu\u00f4ng $ABH$ $\\backsim$ $\\triangle$ vu\u00f4ng $CAH$ $\\Rightarrow$ $AH$ $\\Rightarrow$ $AC$. ","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai16/lv3/img\/H8C3B4_D114.png' \/><\/center><br\/>+) $\\triangle{ABH}$ vu\u00f4ng t\u1ea1i $H$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow$ $\\widehat{B} + \\widehat{BAH} = 90^o$ (hai g\u00f3c ph\u1ee5 nhau)<br\/>L\u1ea1i c\u00f3: $\\widehat{CAH} + \\widehat{BAH} = \\widehat{BAC} = 90^o$ <br\/>$\\Rightarrow \\widehat{B} = \\widehat{CAH}$<br\/>+) X\u00e9t $\\triangle{ABH}$ v\u00e0 $\\triangle{CAH}$ c\u00f3:<br\/>$\\widehat{B} = \\widehat{CAH}$ (ch\u1ee9ng minh tr\u00ean)<br\/>$\\widehat{AHC} = \\widehat{BHA}$ (c\u00f9ng = $90^o$)<br\/>$\\Rightarrow$ $\\triangle{ABH}$ $\\backsim$ $\\triangle{CAH}$ (g\u00f3c - g\u00f3c)<br\/>$\\Rightarrow$ $\\dfrac{AH}{CH} = \\dfrac{BH}{AH}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7)<br\/>$\\Rightarrow$ $AH^2 = BH.CH = 9.16 = 144 \\Rightarrow AH = 12 \\text{(cm)}$<br\/>+) $\\triangle{ACH}$ vu\u00f4ng t\u1ea1i $H$ (gi\u1ea3 thi\u1ebft)<br\/> $\\Rightarrow AH^2 + CH^2 = AC^2$ (\u0111\u1ecbnh l\u00ed py-ta-go)<br\/> $\\Rightarrow AC^2 = 16^2 + 12^2 = 400 \\Rightarrow AC = 20 \\text{(cm)}$<br\/>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 <span class='basic_pink'>$20$<\/span> "}]}],"id_ques":1732},{"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":[[["30"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{BAC} = 100^o$, $\\dfrac{AB}{BC} = \\dfrac{BC}{AB + AC}$. T\u00ednh $\\widehat{ABC}$<br\/><b>\u0110\u00e1p \u00e1n:<\/b> $\\widehat{B} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^o$<\/span>","hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai16/lv3/img\/H8C3B3_K102.png' \/><\/center><br\/>+) V\u1ebd $AD$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BAC}$, $D \\in BC$<br\/>$\\Rightarrow$ $\\dfrac{BD}{DC} = \\dfrac{AB}{AC}$ (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c)<br\/>$\\Rightarrow$ $\\dfrac{BD}{AB} = \\dfrac{DC}{AC}$<br\/>\u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau, ta c\u00f3:<br\/>$\\dfrac{BD}{AB} = \\dfrac{DC}{AC} = \\dfrac{BD + DC}{AB + AC} = \\dfrac{BC}{AB + AC}$ <br\/>M\u00e0 $\\dfrac{AB}{BC} = \\dfrac{BC}{AB + AC}$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow$ $\\dfrac{AB}{BC} = \\dfrac{DB}{AB}$ <br\/>+) X\u00e9t $\\triangle{BAD}$ v\u00e0 $\\triangle{BCA}$ c\u00f3:<br\/>$\\widehat{B}$ chung<br\/>$\\dfrac{AB}{BC} = \\dfrac{DB}{AB}$ (ch\u1ee9ng minh tr\u00ean) <br\/>$\\Rightarrow$ $\\triangle{BAD}$ $\\backsim$ $\\triangle{BCA}$ (c\u1ea1nh-g\u00f3c-c\u1ea1nh)<br\/>$\\Rightarrow$ $\\widehat{BAD} = \\widehat{BCA}$ (c\u1eb7p g\u00f3c t\u01b0\u01a1ng \u1ee9ng)<br\/> L\u1ea1i c\u00f3: $\\widehat{BAD} = \\dfrac{\\widehat{BAC}}{2} = \\dfrac{\\widehat{100}}{2} = 50^o$ (v\u00ec $AD$ l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BAC}$)<br\/>$\\Rightarrow \\widehat{BCA} = 50^o$<br\/> $\\triangle{ABC}$ c\u00f3: $\\widehat{ABC} + \\widehat{BCA} + \\widehat{BAC} = 180^o$ (\u0111\u1ecbnh l\u00ed t\u1ed5ng ba g\u00f3c trong m\u1ed9t tam gi\u00e1c) <br\/>$\\Rightarrow \\widehat{ABC} = 180^o - (\\widehat{BAC} + \\widehat{BCA}) = 180^o - (100^o + 50^o) = 30^o$<br\/>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 <span class='basic_pink'>$30$<\/span><br\/><span class='basic_left'><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/> Khi b\u00e0i to\u00e1n cho c\u00e1c t\u1ec9 s\u1ed1 b\u1eb1ng nhau v\u00e0 c\u00f3 xu\u1ea5t hi\u1ec7n d\u1ea1ng t\u1ec9 s\u1ed1 $\\dfrac{BC}{AB + AC}$ th\u00ec ch\u00fang ta c\u1ea7n ngh\u0129 \u0111\u1ebfn vi\u1ec7c v\u1ebd th\u00eam $AD$ l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BAC}$<\/span> <br\/><br\/> "}]}],"id_ques":1733},{"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":" <span class='basic_left'> Cho tam gi\u00e1c $ABC$, $M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh $BC$ v\u00e0 c\u1ea1nh $BC$ c\u00f3 \u0111\u1ed9 d\u00e0i b\u1eb1ng $a$. Bi\u1ebft $\\widehat{MAB} = \\widehat{C}$. T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh $AB$ theo $a$.<br\/><\/span>","select":[" A. $AB = \\dfrac{a}{2}$"," B. $AB = \\dfrac{3a}{2}$","C. $AB = \\dfrac{a \\sqrt{2}}{2}$","D. $AB = \\dfrac{a \\sqrt{3}}{2}$"],"hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai16/lv3/img\/H8C3B3_K103.png' \/><\/center><br\/>X\u00e9t $\\triangle{ABM}$ v\u00e0 $\\triangle{CBA}$ c\u00f3:<br\/>$\\widehat{B}$ chung<br\/> $\\widehat{MAB} = \\widehat{C}$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow$ $\\triangle{ABM}$ $\\backsim$ $\\triangle{CBA}$ (g\u00f3c - g\u00f3c)<br\/>$\\Rightarrow$ $\\dfrac{AB}{BC} = \\dfrac{BM}{BA}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7) $\\Rightarrow AB^2 = BC.BM$<br\/>L\u1ea1i c\u00f3: $M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow BM = \\dfrac{1}{2}.BC = \\dfrac{a}{2}$<br\/>$\\Rightarrow AB^2 = a. \\dfrac{a}{2} = \\dfrac{a^2}{2} \\Rightarrow AB = \\dfrac{a \\sqrt{2}}{2}$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>C. $AB = \\dfrac{a \\sqrt{2}}{2}$<\/span> ","column":2}]}],"id_ques":1734},{"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":10,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$, \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $CD$. So s\u00e1nh $CD^2$ v\u00e0 $CA.CB$.<br\/><b>\u0110\u00e1p \u00e1n:<\/b> $CD^2$ _input_ $CA.CB$<\/span>","hint":"V\u1ebd th\u00eam h\u00ecnh: Tr\u00ean $BC$ l\u1ea5y \u0111i\u1ec3m $F$ sao cho $\\widehat{CDF} = \\widehat{CAB}$.","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai16/lv3/img\/H8C3B3_K104.png' \/><\/center><br\/>Tr\u00ean $BC$ l\u1ea5y \u0111i\u1ec3m $F$ sao cho $\\widehat{CDF} = \\widehat{CAB}$<br\/>Ta c\u00f3: $CD$ l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c $ABC$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow$ $\\widehat{C_1} = \\widehat{C_2}$ <br\/>X\u00e9t $\\triangle{CAD}$ v\u00e0 $\\triangle{CDF}$ c\u00f3:<br\/>$\\widehat{C_1} = \\widehat{C_2}$ (ch\u1ee9ng minh tr\u00ean)<br\/> $\\widehat{CAD} = \\widehat{CDF}$ (c\u00e1ch v\u1ebd)<br\/>$\\Rightarrow$ $\\triangle{CAD}$ $\\backsim$ $\\triangle{CDF}$ (g\u00f3c - g\u00f3c)<br\/> $\\Rightarrow$ $\\dfrac{CA}{CD} = \\dfrac{CD}{CF}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7)<br\/>$\\Rightarrow$ $CD^2 = CA.CF$<br\/>L\u1ea1i c\u00f3: $CF < CB$ <br\/>$\\Rightarrow CA. CF < CA.CB$ hay $CD^2 < CA.CB$<br\/>V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 <span class='basic_pink'>\"<\"<\/span><br\/><span class='basic_left'><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/> Ngo\u00e0i c\u00e1ch l\u00e0m nh\u01b0 tr\u00ean ch\u00fang ta c\u00f3 th\u1ec3 l\u00e0m nh\u01b0 sau:<br\/>+ V\u00ec $\\widehat{CDB} > \\widehat{CAB}$ n\u00ean tr\u00ean tia \u0111\u1ed1i c\u1ee7a $DC$ ta l\u1ea5y \u0111i\u1ec3m $E$ sao cho $\\widehat{CAE} = \\widehat{CDB}$<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai16/lv3/img\/H8C3B3_K105.png' \/><\/center><br\/>T\u1eeb \u0111\u00f3 ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 nh\u01b0 tr\u00ean \u0111\u1ec3 ra \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3.<\/span> <br\/> "}]}],"id_ques":1735},{"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":"<span class='basic_left'> Cho tam gi\u00e1c nh\u1ecdn $ABC$, c\u00e1c \u0111\u01b0\u1eddng cao $AD, BE, CF$.<br\/><b>H\u00e3y ch\u1ecdn nh\u1eefng \u0111\u00e1p \u00e1n \u0111\u00fang<\/b><\/span><br\/>","hint":"X\u00e9t c\u00e1c c\u1eb7p tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng.","column":2,"number_true":2,"select":["A. $AE.AC = AF.AB$","B. $\\triangle{AEF} \\backsim \\triangle{ABC} $","C. $\\widehat{AEF} = \\widehat{ABC}$","D. $AE.BC = AF.AB$"],"explain":"<span class='basic_left'><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai16/lv3/img\/H8C3B3_K106.png' \/><\/center><br\/> X\u00e9t $\\triangle{ABE}$ v\u00e0 $\\triangle{ACF}$ c\u00f3:<br\/>$\\widehat{AEB} = \\widehat{AFC}$ (c\u00f9ng $= 90^o$)<br\/> $\\widehat{BAC}$ chung<br\/>$\\Rightarrow$ $\\triangle{ABE}$ $\\backsim$ $\\triangle{ACF}$ (g\u00f3c - g\u00f3c)<br\/>$\\Rightarrow$ $\\dfrac{AB}{AC} = \\dfrac{AE}{AF}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7)<br\/>$ \\Rightarrow AE.AC = AF.AB$ <b>(\u0111\u00e1p \u00e1n A \u0111\u00fang, \u0111\u00e1p \u00e1n D sai)<\/b><br\/>$\\Rightarrow$ $\\dfrac{AE}{AB} = \\dfrac{AF}{AC}$<br\/>X\u00e9t $\\triangle{AEF}$ v\u00e0 $\\triangle{ABC}$ c\u00f3:<br\/>$\\dfrac{AE}{AB} = \\dfrac{AF}{AC}$ (ch\u1ee9ng minh tr\u00ean)<br\/> $\\widehat{BAC}$ chung<br\/>$\\Rightarrow$ $\\triangle{AEF}$ $\\backsim$ $\\triangle{ABC}$ (c\u1ea1nh-g\u00f3c-c\u1ea1nh ) <b>(\u0111\u00e1p \u00e1n B \u0111\u00fang)<\/b><br\/> $\\Rightarrow$ $\\widehat{AEF} = \\widehat{ABC}$ (c\u1eb7p g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <b>(\u0111\u00e1p \u00e1n C \u0111\u00fang)<\/b><br\/>V\u1eady nh\u1eefng \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>A, B, C<\/span><br\/><span class='basic_left'><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>\u0110\u1ec3 ch\u1ee9ng minh \u0111\u1eb3ng th\u1ee9c t\u00edch, th\u00f4ng th\u01b0\u1eddng ch\u00fang ta bi\u1ebfn \u0111\u1ed5i ch\u00fang d\u01b0\u1edbi d\u1ea1ng t\u1ec9 l\u1ec7 th\u1ee9c v\u00e0 ch\u1ee9ng minh t\u1ec9 l\u1ec7 th\u1ee9c \u1ea5y.<\/span> <br\/><br\/> "}]}],"id_ques":1736},{"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":" <span class='basic_left'> Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{ABC} = 2\\widehat{ACB}$, $AB = 6cm$, $AC = 9cm$. T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh $BC$.<br\/><\/span>","select":[" A. $BC = 3cm$"," B. $BC = 5cm$","C. $BC = 6cm$","D. $BC = 7,5cm$"],"hint":"K\u1ebb \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $BD$ c\u1ee7a tam gi\u00e1c $ABC$","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai16/lv3/img\/H8C3B3_K107.png' \/><\/center><br\/>K\u1ebb \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $BD$ c\u1ee7a tam gi\u00e1c $ABC$<br\/>$\\Rightarrow$ $\\widehat{ABD} = \\widehat{CBD} = \\dfrac{\\widehat{ABC}}{2}$ (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c)<br\/>L\u1ea1i c\u00f3: $\\widehat{ABC} = 2\\widehat{ACB}$ (gi\u1ea3 thi\u1ebft) $\\Rightarrow$ $\\widehat{ACB}= \\dfrac{\\widehat{ABC}}{2}$ <br\/> $\\Rightarrow$ $\\widehat{ABD} = \\widehat{ACB}$<br\/>X\u00e9t $\\triangle{ABC}$ v\u00e0 $\\triangle{ADB}$ c\u00f3:<br\/>$\\widehat{A}$ chung<br\/> $\\widehat{ABD} = \\widehat{ACB}$ (ch\u1ee9ng minh tr\u00ean)<br\/>$\\Rightarrow$ $\\triangle{ABC}$ $\\backsim$ $\\triangle{ADB}$ (g\u00f3c - g\u00f3c)<br\/>$\\Rightarrow$ $\\dfrac{AB}{AD} = \\dfrac{AC}{AB}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7)<br\/> $\\Rightarrow AD = \\dfrac{AB^2}{AC} = \\dfrac{6^2}{9} = 4 \\text{(cm)}$<br\/>L\u1ea1i c\u00f3: $AD + DC = AC \\Rightarrow DC = AC - AD = 9 - 4 = 5 \\text{(cm)}$<br\/>$\\triangle{ABC}$ c\u00f3: $BD$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c (c\u00e1ch v\u1ebd)<br\/>$\\Rightarrow$ $\\dfrac{BC}{AB} = \\dfrac{CD}{AD}$ (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c)<br\/>$\\Rightarrow BC = \\dfrac{AB.CD}{AD} = \\dfrac{6.5}{4} = 7,5 \\text{(cm)}$ <br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>D. $BC = 7,5cm$<\/span><br\/><span class='basic_left'><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/> T\u1eeb d\u1eef ki\u1ec7n $\\widehat{ABC} = 2\\widehat{ACB}$ $\\Rightarrow$ k\u1ebb \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $BD$ c\u1ee7a tam gi\u00e1c $ABC$ \u0111\u1ec3 t\u1ea1o ra c\u00e1c c\u1eb7p g\u00f3c b\u1eb1ng nhau $\\Rightarrow$ C\u00e1c c\u1eb7p tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng $\\Rightarrow$ C\u00e1c \u0111o\u1ea1n th\u1eb3ng t\u1ec9 l\u1ec7 $\\Rightarrow$ T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh c\u1ea7n t\u00ecm.<\/span> ","column":2}]}],"id_ques":1737},{"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":" <span class='basic_left'> Cho $\\triangle{ABC}$ $\\backsim$ $\\triangle{HIK}$ theo t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng $k = \\dfrac{3}{5}$. T\u00ednh chu vi $\\triangle{HIK}$ bi\u1ebft chu vi $\\triangle{ABC}$ b\u1eb1ng $60cm$.<br\/><\/span>","select":[" A. $C_{\\triangle{HIK}} = 40cm$"," B. $C_{\\triangle{HIK}} = 80cm$","C. $C_{\\triangle{HIK}} = 100cm$","D. $C_{\\triangle{HIK}} = 120cm$"],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\triangle{ABC}$ $\\backsim$ $\\triangle{HIK}$ theo t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng $k = \\dfrac{3}{5}$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow \\begin{cases} AB = \\dfrac{3}{5}. HI \\\\ AC = \\dfrac{3}{5}. HK \\\\ BC = \\dfrac{3}{5}. IK \\end{cases}$ <br\/> $\\Rightarrow AB + AC + BC = \\dfrac{3}{5} (HI + HK + IK)$ <br\/>Hay $C_{\\triangle{ABC}} = \\dfrac{3}{5}.C_{\\triangle{HIK}}$ <br\/>$\\Rightarrow C_{\\triangle{HIK}} = \\dfrac{5}{3} C_{\\triangle{ABC}} = \\dfrac{5}{3}. 60 = 100 \\text{(cm)}$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>C. $C_{\\triangle{HIK}} = 100cm$<\/span><br\/><span class='basic_left'><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>Hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng c\u00f3 t\u1ec9 s\u1ed1 chu vi b\u1eb1ng t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng.<\/span> <br\/><br\/> ","column":2}]}],"id_ques":1738},{"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":" <span class='basic_left'> Cho $\\triangle{ABC}$ v\u00e0 c\u00e1c \u0111\u01b0\u1eddng cao $BD$, $CE$. T\u00ednh $\\widehat{AED}$ bi\u1ebft $\\widehat{ACB} = 50^o$.<br\/><\/span>","select":[" A. $\\widehat{AED} = 25^o$"," B. $\\widehat{AED} = 50^o$","C. $\\widehat{AED} = 100^o$","D. $\\widehat{AED} = 125^o$"],"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 $\\triangle{ABD}$ $\\backsim$ $\\triangle{ACE}$<br\/><b>B\u01b0\u1edbc 2:<\/b> Ch\u1ee9ng minh $\\triangle{ADE}$ $\\backsim$ $\\triangle{ABC}$<\/span><br\/> <span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai16/lv3/img\/H8C3B3_K108.png' \/><\/center><br\/> +) $BD$, $CE$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a $\\triangle{ABC}$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow \\widehat{AEC} = \\widehat{ADB} = 90^o$ <br\/>+) X\u00e9t $\\triangle{ABD}$ v\u00e0 $\\triangle{ACE}$ c\u00f3: <br\/>$\\widehat{AEC} = \\widehat{ADB}$ (ch\u1ee9ng minh tr\u00ean)<br\/> $\\widehat{A} $ chung<br\/>$\\Rightarrow \\triangle{AEC}$ $\\backsim$ $\\triangle{ADB}$ (g\u00f3c - g\u00f3c)<br\/>$\\Rightarrow \\dfrac{AD}{AE} = \\dfrac{AB}{AC}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7)<br\/> +) X\u00e9t $\\triangle{ADE}$ v\u00e0 $\\triangle{ABC}$ c\u00f3: <br\/>$\\widehat{A} $ chung<br\/> $\\dfrac{AD}{AE} = \\dfrac{AB}{AC}$ (ch\u1ee9ng minh tr\u00ean) <br\/>$\\Rightarrow \\triangle{ADE}$ $\\backsim$ $\\triangle{ABC}$ (c\u1ea1nh-g\u00f3c-c\u1ea1nh)<br\/>$\\Rightarrow \\widehat{AED} = \\widehat{ACB}$ (c\u1eb7p g\u00f3c t\u01b0\u01a1ng \u1ee9ng)<br\/> M\u00e0 $\\widehat{ACB} = 50^o$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow \\widehat{AED} = 50^o$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0<span class='basic_pink'> B. $\\widehat{AED} = 50^o$<\/span> <\/span> <br\/><\/span> ","column":2}]}],"id_ques":1739}],"lesson":{"save":0,"level":3}}