{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":" T\u00ecm c\u00e2u tr\u1ea3 l\u1eddi sai. ","select":["A. G\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c l\u00e0 g\u00f3c k\u1ec1 b\u00f9 v\u1edbi m\u1ed9t g\u00f3c c\u1ee7a tam gi\u00e1c.","B. M\u1ed7i g\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c b\u1eb1ng t\u1ed5ng hai g\u00f3c trong kh\u00f4ng k\u1ec1 v\u1edbi n\u00f3.","C. G\u00f3c ngo\u00e0i c\u1ee7a m\u1ed9t tam gi\u00e1c l\u1edbn h\u01a1n m\u1ed7i g\u00f3c trong kh\u00f4ng k\u1ec1 v\u1edbi n\u00f3.","D. T\u1ed5ng c\u00e1c g\u00f3c ngo\u00e0i t\u1ea1i ba \u0111\u1ec9nh c\u1ee7a m\u1ed9t tam gi\u00e1c b\u1eb1ng $ 180^{o}$ "],"explain":" <span class='basic_left'> <span class='basic_pink'> \u0110\u00e1p \u00e1n sai l\u00e0 D. V\u00ec: <\/span><br\/> G\u1ecdi $\\widehat{A_{1}}, \\widehat{B_{1}}, \\widehat{C_{1}} $ theo th\u1ee9 t\u1ef1 l\u00e0 c\u00e1c g\u00f3c ngo\u00e0i t\u1ea1i c\u00e1c \u0111\u1ec9nh A, B, C c\u1ee7a tam gi\u00e1c ABC. <br\/> Ta c\u00f3 $\\widehat{A_{1}} + \\widehat{A} = 180^{o}, \\widehat{B_{1}} + \\widehat{B} = 180^{o}, \\widehat{C_{1}} + \\widehat{C} = 180^{o} \\\\ \\Rightarrow (\\widehat{A_{1}} + \\widehat{B_{1}} + \\widehat{C_{1}}) + (\\widehat{A} + \\widehat{B} + \\widehat{C} ) = 180^{o} + 180^{o} + 180^{o} = 540^{o}$ <br\/> m\u00e0 $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o} $ <br\/> Suy ra $\\widehat{A_{1}} + \\widehat{B_{1}} + \\widehat{C_{1}} = 540^{o} - 180^{o} = 360^{o} $ <\/span> ","column":1}]}],"id_ques":1571},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["60"],["40"],["80"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> T\u00ednh c\u00e1c g\u00f3c c\u1ee7a tam gi\u00e1c $ABC$ bi\u1ebft: $4\\widehat{A} = 6\\widehat{B} = 3\\widehat{C} $ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> <br\/> $\\widehat{A} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} \\\\ \\widehat{B} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} \\\\ \\widehat{C} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} $ ","explain":" <span class='basic_left'>X\u00e9t tam gi\u00e1c ABC c\u00f3: $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o} $ (1) <br\/> Ta c\u00f3: $4\\widehat{A} = 6\\widehat{B} = 3\\widehat{C} $ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow \\dfrac{4\\widehat{A}}{12} = \\dfrac{6\\widehat{B}}{12} = \\dfrac{3\\widehat{C}}{12} \\Rightarrow \\dfrac{\\widehat{A}}{3} = \\dfrac{\\widehat{B}}{2} = \\dfrac{\\widehat{C}}{4} $ (2) <br\/> T\u1eeb (1) v\u00e0 (2), \u00e1p d\u1ee5ng t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau ta c\u00f3: <br\/> $ \\dfrac{\\widehat{A}}{3} = \\dfrac{\\widehat{B}}{2} = \\dfrac{\\widehat{C}}{4} = \\dfrac{\\widehat{A}+\\widehat{B}+\\widehat{C}}{3+2+4} = \\dfrac{180^{o}}{9} = 20^{o} $ <br\/> Suy ra $\\widehat{A} = 20^{o}.3 = 60^{o} \\\\ \\widehat{B} = 20^{o}.2 = 40^{o} \\\\ \\widehat{C} = 20^{o}.4 = 80^{o} $ <br\/> <br\/> <span class='basic_pink'> V\u1eady $\\widehat{A} = 60^{o} \\\\ \\widehat{B} = 40^{o} \\\\ \\widehat{C} = 80^{o} $<\/span><\/span> <br\/>"}]}],"id_ques":1572},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["45"],["75"],["60"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"T\u00ednh c\u00e1c g\u00f3c c\u1ee7a tam gi\u00e1c $MNP$ bi\u1ebft: $5\\widehat{M} = 3\\widehat{N}, 7\\widehat{M} - 4\\widehat{N} = 15^{o} $ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> <br\/> $\\widehat{M} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} \\\\ \\widehat{N} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} \\\\ \\widehat{P} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} $ ","explain":" <span class='basic_left'>Ta c\u00f3: $5\\widehat{M} = 3\\widehat{N}$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow \\dfrac{\\widehat{M}}{3} = \\dfrac{\\widehat{N}}{5} $ <br\/> \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau, ta c\u00f3: <br\/> $\\dfrac{\\widehat{M}}{3} = \\dfrac{\\widehat{N}}{5} = \\dfrac{7\\widehat{M}}{21} = \\dfrac{4\\widehat{N}}{20} = \\dfrac{7\\widehat{M} - 4\\widehat{N} }{21-20} = \\dfrac{15^{o}}{1} = 15^{o} $ <br\/> Suy ra $\\widehat{M} = 15^{o}.3 = 45^{o} \\\\ \\widehat{N} = 15^{o}.5 = 75^{o} $ <br\/> X\u00e9t tam gi\u00e1c $MNP$ c\u00f3: $\\widehat{M} + \\widehat{N} + \\widehat{P} = 180^{o} $ <br\/> M\u00e0 $ \\widehat{M} = 45^{o} , \\widehat{N} = 75^{o} $ <br\/> N\u00ean $\\widehat{P} = 180^{o} - (45^{o} + 75^{o} ) = 60^{o} $ <br\/> <span class='basic_pink'> V\u1eady $\\widehat{M} = 45^{o} \\\\ \\widehat{N} = 75^{o} \\\\ \\widehat{P} = 60^{o} $<\/span><\/span> <br\/>"}]}],"id_ques":1573},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"B\u00e0i n\u00e0y g\u1ed3m 2 c\u00e2u h\u1ecfi","temp":"fill_the_blank","correct":[[["140"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{A} = 100^{o} ,$ hai tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{B} $ v\u00e0 $\\widehat{C}$ c\u1eaft nhau t\u1ea1i I. G\u1ecdi $Cx$ l\u00e0 tia \u0111\u1ed1i c\u1ee7a tia $CB$, hai tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{B}$ v\u00e0 $\\widehat{ACx}$ c\u1eaft nhau t\u1ea1i N. <br\/> <b> C\u00e2u 1: <\/b> S\u1ed1 \u0111o g\u00f3c $\\widehat{BIC} $ l\u00e0: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} $<\/span>","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv3/img\/H7B9-24.png' \/> <\/center> <span class='basic_left'> X\u00e9t tam gi\u00e1c $BIC$, ta c\u00f3: <br\/> $\\widehat{BIC} + \\widehat{B_{1}} + \\widehat{C_{1}} = 180^{o} \\\\ \\Rightarrow \\widehat{BIC} = 180^{o} - (\\widehat{B_{1}} + \\widehat{C_{1}}) \\,\\,\\,(1)$ <br\/> X\u00e9t tam gi\u00e1c $ABC$, ta c\u00f3: $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o} $ m\u00e0 $\\widehat{A} = 100^{o} $ <br\/> $\\Rightarrow \\widehat{B} + \\widehat{C} = 180^{o} - 100^{o} = 80^{o} $ <br\/> V\u00ec BI v\u00e0 CI l\u1ea7n l\u01b0\u1ee3t l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{B} $ v\u00e0 $\\widehat{C} $ <br\/> $ \\widehat{B_{1}} + \\widehat{C_{1}} = \\dfrac{\\widehat{B}}{2} + \\dfrac{\\widehat{C}}{2} = \\dfrac{\\widehat{B}+\\widehat{C}}{2} = \\dfrac{80^{o}}{2} = 40^{o} $ (2) <br\/> T\u1eeb (1) v\u00e0 (2) suy ra: $\\widehat{BIC} = 180^{o} - 40^{o} = 140^{o}$ <br\/> <br\/> <span class='basic_pink'> V\u1eady $\\widehat{BIC} = 140^{o} $<\/span><\/span> <br\/>"}]}],"id_ques":1574},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"B\u00e0i n\u00e0y g\u1ed3m 2 c\u00e2u h\u1ecfi","temp":"fill_the_blank","correct":[[["50"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{A} = 100^{o} ,$ hai tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{B} $ v\u00e0 $\\widehat{C}$ c\u1eaft nhau t\u1ea1i I. G\u1ecdi $Cx$ l\u00e0 tia \u0111\u1ed1i c\u1ee7a tia $CB$, hai tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{B}$ v\u00e0 $\\widehat{ACx}$ c\u1eaft nhau t\u1ea1i N. <br\/> <b> C\u00e2u 2: <\/b> S\u1ed1 \u0111o g\u00f3c $\\widehat{BNC} $ l\u00e0: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} $<\/span>","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv3/img\/H7B9-24.png' \/> <\/center> <span class='basic_left'>V\u00ec $CI$ v\u00e0 $CN$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 hai tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{ACB} $ v\u00e0 $\\widehat{ACx} $ <br\/> $\\Rightarrow \\widehat{ICN} = \\widehat{C_{2}} + \\widehat{C_{3}} = \\dfrac{\\widehat {ACB}}{2} + \\dfrac{\\widehat{ACx}}{2} $<br\/>$= \\dfrac{\\widehat{ACB}+\\widehat{ACx}}{2} = \\dfrac{180^{o}}{2} = 90^{o} $ <br\/> X\u00e9t tam gi\u00e1c $NIC$ c\u00f3 $\\widehat{BIC} $ l\u00e0 g\u00f3c ngo\u00e0i n\u00ean: <br\/> $\\widehat{BIC} = \\widehat{ICN} + \\widehat{BNC} \\\\ \\Leftrightarrow 140^{o} = 90^{o} + \\widehat{BNC} \\\\ \\Rightarrow \\widehat{BNC} = 140^{o} - 90^{o} = 50^{o} $ <br\/> <br\/> <span class='basic_pink'> V\u1eady $\\widehat{BNC} = 50^{o} $<\/span><\/span> <br\/>"}]}],"id_ques":1575},{"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":"Gi\u00e1 tr\u1ecb c\u1ee7a x \u1edf h\u00ecnh sau l\u00e0: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv3/img\/H7B9-25.png' \/> <\/center> ","select":[" A. $105^{o} $ "," B. $110^{o} $ "," C. $120^{o} $ ","D. $115^{o} $ "],"hint":"Ta \u0111i t\u00ednh g\u00f3c $ACB$ r\u1ed3i suy ra s\u1ed1 \u0111o g\u00f3c $ECD$. <br\/> x l\u00e0 s\u1ed1 \u0111o c\u1ee7a g\u00f3c ngo\u00e0i t\u1ea1i \u0111\u1ec9nh E c\u1ee7a $\\Delta CDE$","explain":" <span class='basic_left'> Tam gi\u00e1c $ABC$ ta c\u00f3: $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o}$ <br\/> n\u00ean $83^{o} + 45^{o} + \\widehat{C} = 180^{o}$ <br\/> $\\Rightarrow \\widehat{C} = 52^{o} $ <br\/> M\u1eb7t kh\u00e1c $\\widehat{ACB} = \\widehat{ECD} = 52^{o} $ (\u0111\u1ed1i \u0111\u1ec9nh) <br\/> Ta l\u1ea1i c\u00f3: x l\u00e0 s\u1ed1 \u0111o c\u1ee7a g\u00f3c ngo\u00e0i t\u1ea1i \u0111\u1ec9nh E c\u1ee7a tam gi\u00e1c $ECD$: <br\/> $\\Rightarrow x = 52^{o} + 53^{o} = 105^{o}$ <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 A. <\/span>","column":4}]}],"id_ques":1576},{"time":24,"part":[{"time":3,"title":"N\u1ed1i t\u1eeb ho\u1eb7c c\u1ee5m t\u1eeb \u1edf c\u1ed9t b\u00ean tr\u00e1i v\u1edbi c\u1ed9t b\u00ean ph\u1ea3i \u0111\u1ec3 \u0111\u01b0\u1ee3c c\u00e2u ho\u00e0n ch\u1ec9nh","title_trans":"","audio":"","temp":"matching","correct":[["3","2","1","4"]],"list":[{"point":10,"image":"https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv3/img\/H7B9-27.png","left":["Trong tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o} $","$\\widehat{ACy} $ l\u00e0 g\u00f3c ngo\u00e0i t\u1ea1i \u0111\u1ec9nh C c\u1ee7a tam gi\u00e1c ABC","V\u00ec $Cx$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{ACy} $","C\u00f3 c\u1eb7p g\u00f3c so le trong $\\widehat{A} = \\widehat{C_{1}} = 60^{o}$"],"right":["n\u00ean $\\widehat{C_{1}} = \\widehat{C_{2}} $$= \\dfrac{1}{2}\\widehat{ACy} = 60^{o}$","n\u00ean $\\widehat{ACy} = 60^{o} + 60^{o} = 120^{o} $","n\u00ean $\\widehat{C} = 180^o-(60^o+60^o)$$=60^{o} $","do \u0111\u00f3 $AB \/\/ Cx$"],"top":100,"explain":" <span class='basic_left'> Trong tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o} $ <br\/> n\u00ean $\\widehat{C} = 180^{o} - (60^{o} + 60^{o}) = 60^{o} $ <br\/> $\\widehat{ACy} $ l\u00e0 g\u00f3c ngo\u00e0i t\u1ea1i \u0111\u1ec9nh C c\u1ee7a tam gi\u00e1c $ABC$ <br\/> n\u00ean $\\widehat{ACy} = 60^{o} + 60^{o} = 120^{o} $ <br\/> V\u00ec Cx l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{ACy} $ <br\/> n\u00ean $\\widehat{C_{1}} = \\widehat{C_{2}} = \\dfrac{1}{2}\\widehat{ACy} = 60^{o} $ <br\/> Do \u0111\u00f3: $\\widehat{A} = \\widehat{C_{1}} = 60^{o} $ <br\/> M\u00e0 hai g\u00f3c $\\widehat{C_1}$ v\u00e0 $\\widehat {A}$ \u1edf v\u1ecb tr\u00ed so le trong v\u1edbi nhau <br\/> $\\Rightarrow AB \/\/ Cx$ "}]}],"id_ques":1577},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["70"],["20"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{B} > \\widehat{C} $. K\u1ebb $AH$ vu\u00f4ng g\u00f3c v\u1edbi $BC$ t\u1ea1i H, k\u1ebb \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $AD$ c\u1ee7a g\u00f3c $\\widehat{A} $ (D thu\u1ed9c BC). Bi\u1ebft $\\widehat{HAD} = 25^{o} $ v\u00e0 $\\widehat{A} = 90^{o} $ <br\/> T\u00ednh s\u1ed1 \u0111o g\u00f3c B v\u00e0 g\u00f3c C? <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> <br\/> $\\widehat{B} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} \\\\ \\widehat{C} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} $ <\/span> ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh s\u1ed1 \u0111\u00f3 g\u00f3c $\\widehat{ADH}$ d\u1ef1a v\u00e0o t\u1ed5ng 3 g\u00f3c trong tam gi\u00e1c $ADH$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh s\u1ed1 \u0111o g\u00f3c $\\widehat{BAD}$=? <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh s\u1ed1 \u0111o g\u00f3c $\\widehat{B}$ d\u1ef1a v\u00e0o t\u1ed5ng 3 g\u00f3c trong tam gi\u00e1c $ADB$ <br\/> <b> B\u01b0\u1edbc 4: <\/b> T\u00ednh s\u1ed1 \u0111o g\u00f3c $\\widehat{C}$ d\u1ef1a v\u00e0o t\u1ed5ng 3 g\u00f3c trong tam gi\u00e1c $ABC$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv3/img\/H7B9-28.png' \/> <\/center> <span class='basic_left'>* Tam gi\u00e1c $AHD$ vu\u00f4ng t\u1ea1i H n\u00ean $\\widehat{HAD} + \\widehat{ADH} = 90^{o} $ <br\/> $\\Rightarrow \\widehat{ADH} = 90^{o} - \\widehat{HAD} = 90^{o} - 25^{o} = 65^{o} $ hay $\\widehat{ADB} = 65^{o} $ <br\/> * $AD$ l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c A n\u00ean $\\widehat{BAD} = \\dfrac{\\widehat{A}}{2} = \\dfrac{90^{o}}{2} = 45^{o} $ <br\/> * Tam gi\u00e1c $ABD$ c\u00f3 $\\widehat{BAD} + \\widehat{B} + \\widehat{ADB} = 180^{o} \\\\ \\Leftrightarrow 45^{o} + \\widehat{B} + 65^{o} = 180^{o} \\\\ \\Leftrightarrow \\widehat{B} = 70^{o} $ <br\/> * Tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i A n\u00ean $\\widehat{B} + \\widehat{C} = 90^{o} $ <br\/> M\u00e0 $\\widehat{B} = 70^{o} $ n\u00ean $\\widehat{C} = 90^{o} - 70^{o} = 20^{o}$ <br\/> <span class='basic_pink'> V\u1eady $\\widehat{B} = 70^{o} \\\\ \\widehat{C} = 20^{o} $<\/span><\/span> <br\/>"}]}],"id_ques":1578},{"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":" <span class='basic_left'> Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{B} - \\widehat{C} = 20^{o} $. Tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c A c\u1eaft $BC$ t\u1ea1i D. K\u1ebb $AH \\perp BC$. S\u1ed1 \u0111o g\u00f3c $\\widehat{HAD} $ l\u00e0: ","select":[" A. $10^{o} $ "," B. $20^{o} $ "," C. $30^{o} $ ","D. $50^{o} $ "],"explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv3/img\/H7B9-29.png' \/> <\/center> <span class='basic_left'>Tam gi\u00e1c $ABD$ c\u00f3: $\\widehat{B} + \\widehat{ADB} = 180^{o} - \\widehat{A_{1}} $ <br\/> Tam gi\u00e1c $ADC$ c\u00f3 $\\widehat{C} + \\widehat{ADC} = 180^{o} - \\widehat{A_{2}} $ <br\/> M\u00e0 $\\widehat{A_{1}} = \\widehat{A_{2}} \\Rightarrow \\widehat{B} + \\widehat{ADB} = \\widehat{C} + \\widehat{ADC} \\\\ \\Rightarrow \\widehat{ADC} - \\widehat{ADB} = \\widehat{B} - \\widehat{C} = 20^{o} $ <br\/> M\u00e0 $\\widehat{ADC} + \\widehat{ADB} = 180^{o} $ (hai g\u00f3c k\u1ec1 b\u00f9) <br\/> Suy ra $2\\widehat{ADB} = 180^{o} - 20^{o} = 160^{o} \\\\ \\Rightarrow \\widehat{ADB} = 80^{o} $ <br\/> Tam gi\u00e1c $AHD$ vu\u00f4ng t\u1ea1i H $\\Rightarrow \\widehat{HAD} + \\widehat{ADH} = 90^{o} \\\\ \\widehat{HAD} = 90^{o} - \\widehat{ADH} = 90^{o} - 80^{o} = 10^{o} $ <br\/><br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 A. <\/span>","column":4}]}],"id_ques":1579},{"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":" <span class='basic_left'> Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i A, k\u1ebb $AH \\perp BC$. C\u00e1c tia ph\u00e2n gi\u00e1c c\u1ee7a c\u00e1c g\u00f3c $\\widehat{C} $ v\u00e0 $\\widehat{BAH} $ c\u1eaft nhau t\u1ea1i K. Khi \u0111\u00f3 $AK \\perp CK $ <b> \u0111\u00fang <\/b> hay <b> sai <\/b>? ","select":["\u0110\u00fang","Sai"],"explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv3/img\/H7B9-30.png' \/> <\/center> <span class='basic_left'>X\u00e9t tam gi\u00e1c $ABH$ v\u00e0 tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i H v\u00e0 A, ta c\u00f3: <br\/> $\\widehat{BAH} = 90^{o} - \\widehat{B}; \\widehat{ACB} = 90^{o} - \\widehat{B} $ <br\/> n\u00ean $\\widehat{BAH} = \\widehat{ACB} $ <br\/> Ta c\u00f3: $ \\widehat{BAK} = \\dfrac{1}{2}. \\widehat{BAH}; \\widehat{HCK} = \\dfrac{1}{2}. \\widehat{ACB} $ <br\/> n\u00ean $ \\widehat{BAK} = \\widehat{ACK}$ <br\/> X\u00e9t tam gi\u00e1c $AKC$, ta c\u00f3: $ \\widehat{KAC} + \\widehat{ACK} = \\widehat{KAC} + \\widehat{BAK} = 90^{o}$ <br\/> Do \u0111\u00f3 tam gi\u00e1c $AKC$ vu\u00f4ng t\u1ea1i K hay $AK \\perp CK $","column":2}]}],"id_ques":1580}],"lesson":{"save":0,"level":3}}