{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["t","t","t","t","f"]],"list":[{"point":10,"image":"","ques":"","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["N\u1ebfu hai tam gi\u00e1c b\u1eb1ng nhau th\u00ec hai c\u1ea1nh v\u00e0 g\u00f3c xen gi\u1eefa c\u1ee7a tam gi\u00e1c n\u00e0y b\u1eb1ng hai c\u1ea1nh v\u00e0 g\u00f3c xen gi\u1eefa c\u1ee7a tam gi\u00e1c kia.","N\u1ebfu hai c\u1ea1nh v\u00e0 g\u00f3c xen gi\u1eefa c\u1ee7a tam gi\u00e1c n\u00e0y b\u1eb1ng hai c\u1ea1nh v\u00e0 g\u00f3c xen gi\u1eefa c\u1ee7a tam gi\u00e1c kia th\u00ec hai tam gi\u00e1c \u0111\u00f3 b\u1eb1ng nhau.","N\u1ebfu hai c\u1ea1nh v\u00e0 g\u00f3c xen gi\u1eefa c\u1ee7a tam gi\u00e1c n\u00e0y b\u1eb1ng hai c\u1ea1nh v\u00e0 g\u00f3c xen gi\u1eefa c\u1ee7a tam gi\u00e1c kia th\u00ec hai c\u1ea1nh c\u00f2n l\u1ea1i c\u0169ng b\u1eb1ng nhau.","N\u1ebfu hai c\u1ea1nh v\u00e0 g\u00f3c xen gi\u1eefa c\u1ee7a tam gi\u00e1c n\u00e0y b\u1eb1ng hai c\u1ea1nh v\u00e0 g\u00f3c xen gi\u1eefa c\u1ee7a tam gi\u00e1c kia th\u00ec hai c\u1eb7p g\u00f3c c\u00f2n l\u1ea1i c\u0169ng t\u01b0\u01a1ng \u1ee9ng b\u1eb1ng nhau.","N\u1ebfu hai tam gi\u00e1c c\u00f3 hai c\u1eb7p g\u00f3c t\u01b0\u01a1ng \u1ee9ng b\u1eb1ng nhau th\u00ec hai tam gi\u00e1c c\u0169ng c\u00f3 hai c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng b\u1eb1ng nhau. "],"explain":["<span class='basic_left'> 1 - \u0110\u00fang. V\u00ec theo \u0111\u1ecbnh ngh\u0129a hai tam gi\u00e1c b\u1eb1ng nhau. <\/span>","<span class='basic_left'>2 - \u0110\u00fang v\u00ec theo t\u00ednh ch\u1ea5t c\u1ea1nh-g\u00f3c-c\u1ea1nh.<\/span>","<span class='basic_left'>3 - \u0110\u00fang v\u00ec theo t\u00ednh ch\u1ea5t c\u1ea1nh-g\u00f3c-c\u1ea1nh ta c\u00f3 hai tam gi\u00e1c b\u1eb1ng nhau n\u00ean hai c\u1ea1nh c\u00f2n l\u1ea1i b\u1eb1ng nhau.<\/span>","<span class='basic_left'>4 - \u0110\u00fang v\u00ec theo t\u00ednh ch\u1ea5t c\u1ea1nh-g\u00f3c-c\u1ea1nh ta c\u00f3 hai tam gi\u00e1c b\u1eb1ng nhau n\u00ean hai c\u1eb7p g\u00f3c t\u01b0\u01a1ng \u1ee9ng c\u00f2n l\u1ea1i b\u1eb1ng nhau.<\/span>","<span class='basic_left'>5 - Sai, v\u00ed d\u1ee5 hai tam gi\u00e1c $ABC$ v\u00e0 $A'B'C'$ c\u00f3 $\\widehat{B} = \\widehat{B'}, \\widehat{B'} = \\widehat{C} $ nh\u01b0ng kh\u00f4ng c\u00f3 hai c\u1eb7p c\u1ea1nh n\u00e0o b\u1eb1ng nhau. <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai11/lv3/img\/H7B11-41.png' \/><\/center> <\/span> "]}]}],"id_ques":1631},{"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 c\u00e2u tr\u1ea3 l\u1eddi sai. <br\/> <span class='basic_left'>Cho tam gi\u00e1c $ABC (AB < AC)$ tr\u00ean c\u1ea1nh $AC$ l\u1ea5y E sao cho $AE = AB.$ Ta c\u00f3: $DB = DE$ (D $\\in $ BC) khi: <\/span> ","select":[" A. $AD$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{A} $"," B. $AD$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $BE$. "," C. D l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $BC$. ","D. $\\triangle ABD = \\triangle AED $ "],"explain":" <img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai11/lv3/img\/H7B11-12A.png' \/> <img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai11/lv3/img\/H7B11-12B.png' \/> <img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai11/lv3/img\/H7B11-12C.png' \/> <img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai11/lv3/img\/H7B11-12D.png' \/> <span class='basic_left'> A. \u0110\u00fang. V\u00ec: + $AB=AE$ (gi\u1ea3 thi\u1ebft) <br\/> + $\\widehat{BAD}=\\widehat{EAD}$ <br\/> + $AD$ c\u1ea1nh chung <br\/> $\\Rightarrow \\triangle ABD = \\triangle AED $ (c.g.c) suy ra $DB = DE$ <br\/> B. \u0110\u00fang. G\u1ecdi K l\u00e0 giao \u0111i\u1ec3m c\u1ee7a trung tr\u1ef1c $AD$ v\u1edbi $BE$ <br\/> $\\Rightarrow BK=EK$ v\u00e0 $AD\\bot BE=K$ <br\/> V\u00ec: + $BK=EK$ (ch\u1ee9ng minh tr\u00ean) <br\/> + $\\widehat{BKD}=\\widehat{EKD}=90^o$ <br\/> + $KD$ l\u00e0 c\u1ea1nh chung <br\/> $\\Rightarrow \\triangle BKD = \\triangle EKD $ (c.g.c) suy ra $DB = DE$ <br\/> C. Sai. Kh\u00f4ng \u0111\u1ee7 \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 $DB = DE$ <br\/> D. \u0110\u00fang. V\u00ec $\\triangle ABD = \\triangle AED \\Rightarrow DB = DE$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <\/span> <br\/><br\/><span class='basic_pink'>C\u00e2u tr\u1ea3 l\u1eddi sai l\u00e0 C. <\/span> <\/span>","column":1}]}],"id_ques":1632},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":" Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A, M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$. Ta c\u00f3: ","select":[" A. $AC = \\dfrac{1}{2} BC $"," B. $AM > \\dfrac{1}{2} BC $"," C. $AM < \\dfrac{1}{2} BC $ ","D. $AM = \\dfrac{1}{2} BC $ "],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai11/lv3/img\/H7B11-15.png' \/><\/center> <span class='basic_left'>Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia MA l\u1ea5y D sao cho $MD = MA = \\dfrac{DA}{2} $ <br\/> X\u00e9t $\\triangle AMB $ v\u00e0 $\\triangle DMC $ c\u00f3: <br\/> + $MA = MD$ (theo c\u00e1ch v\u1ebd \u0111i\u1ec3m D) <br\/> + $\\widehat{AMB} = \\widehat{DMC} $ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh) <br\/> + $MB = MC$ (v\u00ec M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC) <br\/> Do \u0111\u00f3 $\\triangle AMB = \\triangle DMC $ <br\/> $\\Rightarrow AB = DC, \\widehat{MAB} = \\widehat{MDC} $ <br\/> M\u00e0 $\\widehat{MAB}$ v\u00e0 $\\widehat{MDC}$ so le trong n\u00ean $AB \/\/ CD$ <br\/> V\u00ec $AB \\perp AC $ ($\\triangle ABC$ vu\u00f4ng t\u1ea1i A) $\\Rightarrow CD \\perp AC \\Rightarrow \\widehat{ACD} = 90^{o} $ <br\/> X\u00e9t $\\triangle ABC $ v\u00e0 $\\triangle CDA $ c\u00f3: <br\/> + $AB = DC$ (ch\u1ee9ng minh tr\u00ean) <br\/> + $\\widehat{BAC} = \\widehat{DCA} (= 90^{o}) $ <br\/> + $AC$ c\u1ea1nh chung <br\/> Do \u0111\u00f3 $\\triangle ABC = \\triangle CDA (c.g.c) $ <br\/> $\\Rightarrow BC = DA $ <br\/> Do \u0111\u00f3: $AM = \\dfrac{1}{2} BC $ (v\u00ec $MA = \\dfrac{DA}{2} $)<br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 D. <\/span>","column":2}]}],"id_ques":1633},{"time":24,"part":[{"time":3,"title":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u","title_trans":"Qua trung \u0111i\u1ec3m M c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AB k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng d vu\u00f4ng g\u00f3c v\u1edbi AB. Tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng d l\u1ea5y hai \u0111i\u1ec3m H v\u00e0 K sao cho M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a HK. <br\/> H\u00e3y s\u1eafp x\u1ebfp c\u00e1c \u00fd ch\u1ee9ng minh HK l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{AHB} $. ","temp":"sequence","correct":[[[1],[3],[4],[2],[5]]],"list":[{"point":10,"image":"https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai11/lv3/img\/H7B11-29.png","left":["X\u00e9t $\\triangle AMH $ v\u00e0 $\\triangle BMH $ c\u00f3:","Do \u0111\u00f3 $\\triangle AMH = \\triangle BMH $ (hai c\u1ea1nh g\u00f3c vu\u00f4ng b\u1eb1ng nhau)","Suy ra $\\widehat{AHM} = \\widehat{BHM} $ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) ","$\\widehat{AMH} = \\widehat{BMH} = 90^{o} $, c\u1ea1nh HM chung, MA = MB (gi\u1ea3 thi\u1ebft) ","V\u1eady HK l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{AHB} $ "],"top":55,"explain":"<span class='basic_left'> X\u00e9t $\\triangle AMH $ v\u00e0 $\\triangle BMH $ c\u00f3:<br\/>+ c\u1ea1nh HM chung <br\/> + $\\widehat{AMH} = \\widehat{BMH} = 90^{o} $<br\/> + $MA = MB$ (gi\u1ea3 thi\u1ebft) <br\/> Do \u0111\u00f3 $\\triangle AMH = \\triangle BMH $ (hai c\u1ea1nh g\u00f3c vu\u00f4ng b\u1eb1ng nhau) <br\/> Suy ra $\\widehat{AHM} = \\widehat{BHM} $ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/>V\u1eady HK l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{AHB} $ <\/span> "}]}],"id_ques":1634},{"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 tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{B} = 2\\widehat{C} $. Tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c B c\u1eaft AC t\u1ea1i D. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia BD l\u1ea5y \u0111i\u1ec3m N sao cho $BN = AC$. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $CB$ l\u1ea5y \u0111i\u1ec3m P sao cho $CP = AB$. <br\/> T\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u1ee7a g\u00f3c $ACB$ \u0111\u1ec3 $AN \\perp AP $ <\/span>","select":[" A. $\\widehat{ACB} = 90^{o} $"," B. $\\widehat{ACB} = 45^{o} $"," C. $\\widehat{ACB} = 60^{o} $ ","D. $\\widehat{ACB} = 30^{o} $ "],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai11/lv3/img\/H7B11-17.png' \/><\/center> <span class='basic_left'>Ta c\u00f3 $\\widehat{B_{1}} = \\widehat{B_{2}} = \\dfrac{1}{2}\\widehat{ABC} $, m\u00e0 $\\widehat{B} = 2\\widehat{C} $ n\u00ean $\\widehat{B_{1}} = \\widehat{B_{2}} = \\widehat{C_{1}} $ <br\/> M\u1eb7t kh\u00e1c c\u00f3: $\\widehat{B_{2}} + \\widehat{B_{3}} = 180^{o}; \\widehat{C_{1}} + \\widehat{C_{2}} = 180^{o} \\\\ \\Rightarrow \\widehat{B_{3}} = \\widehat{C_{2}}$ <br\/> X\u00e9t $\\triangle ABN $ v\u00e0 $\\triangle PCA $ c\u00f3: <br\/> + $AB = CP$ (gi\u1ea3 thi\u1ebft) <br\/>+ $ \\widehat{B_{3}} = \\widehat{C_{2}}$ (ch\u1ee9ng minh tr\u00ean) <br\/> + $ BN = CA$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow \\triangle ABN = \\triangle PCA (c.g.c) $ <br\/> C\u00f3 $\\triangle ABN = \\triangle PCA \\Rightarrow \\widehat{A_{1}} = \\widehat{N} $ <br\/> V\u1eady $\\widehat{NAP} = 90^{o} \\Leftrightarrow \\widehat{A_{1}} + \\widehat{A_{2}} + \\widehat{A_{3}} = 90^{o} \\Leftrightarrow \\widehat{N} + \\widehat{A_{2}} + \\widehat{A_{3}} = 90^{o} \\\\ \\Leftrightarrow \\widehat{ADB} = 90^{o}$ <br\/> $\\Rightarrow \\widehat{BDC}=90^o$ <br\/> M\u00e0 $\\widehat{B_1}=\\widehat{C_1}$ (ch\u1ee9ng minh tr\u00ean) <br\/> $ \\Leftrightarrow \\widehat{C_1}= \\widehat{ACB} = 45^{o} $ <\/span> <br\/><br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 B. <\/span>","column":2}]}],"id_ques":1635},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u00e2y l\u00e0 g\u00f3i 2 c\u00e2u h\u1ecfi. ","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho $\\widehat{xOy} = 35^{o} $. Tr\u00ean Ox l\u1ea5y \u0111i\u1ec3m A. Qua A k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi Ox c\u1eaft Oy \u1edf B. Qua B k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi Oy c\u1eaft Ox \u1edf C. Qua C k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi Ox c\u1eaft Oy \u1edf D. <br\/> <b><u> C\u00e2u 1: <\/u> <\/b> C\u00f3 bao nhi\u00eau tam gi\u00e1c vu\u00f4ng trong h\u00ecnh v\u1ebd? <\/span>","select":[" A. 3"," B. 4"," C. 5 ","D. 6"],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai11/lv3/img\/H7B11-30.png' \/><\/center> C\u00f3 5 tam gi\u00e1c vu\u00f4ng trong h\u00ecnh v\u1ebd: <br\/> $\\triangle ABO, \\triangle ABC, \\triangle BOC, \\triangle BCD, \\triangle COD $ <br\/><br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 C. <\/span>","column":4}]}],"id_ques":1636},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u00e2y l\u00e0 g\u00f3i 2 c\u00e2u h\u1ecfi. ","temp":"fill_the_blank","correct":[[["55"],["35"],["55"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Cho $\\widehat{xOy} = 35^{o} $. Tr\u00ean Ox l\u1ea5y \u0111i\u1ec3m A. Qua A k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi Ox c\u1eaft Oy \u1edf B. Qua B k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi Oy c\u1eaft Ox \u1edf C. Qua C k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi Ox c\u1eaft Oy \u1edf D. <br\/> <b><u> C\u00e2u 2: <\/u> <\/b> T\u00ednh $\\widehat{OBA}, \\widehat{ABC}, \\widehat{CDO}$ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> <br\/> $\\widehat{OBA} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} \\\\ \\widehat{ABC} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} \\\\ \\widehat{CDO} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} $ ","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai11/lv3/img\/H7B11-30.png' \/><\/center> <span class='basic_left'> Tam gi\u00e1c $AOB$ vu\u00f4ng t\u1ea1i A n\u00ean:<br\/> $\\widehat{AOB} + \\widehat{OBA} = 90^{o} \\\\ \\Rightarrow \\widehat{OBA} = 90^{o} - \\widehat{AOB} = 90^{o} - 35^{o} = 55^{o} $ <br\/> Tam gi\u00e1c OBC vu\u00f4ng t\u1ea1i B n\u00ean:<br\/> $\\widehat{BOC} + \\widehat{OCB} = 90^{o} \\\\ \\Rightarrow \\widehat{OCB} = 90^{o} - \\widehat{BOC} = 90^{o} - 35^{o} = 55^{o} $ <br\/> Tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A n\u00ean:<br\/> $\\widehat{ACB} + \\widehat{ABC} = 90^{o} \\\\ \\Rightarrow \\widehat{ABC} = 90^{o} - \\widehat{ACB} = 90^{o} - 55^{o} = 35^{o} $ <br\/> V\u00ec $AB \\perp Ox, CD \\perp Ox $ suy ra AB \/\/ CD do \u0111\u00f3 $\\widehat{CDO} = \\widehat{OBA} = 55^{o} $ (hai g\u00f3c \u0111\u1ed3ng v\u1ecb) <\/span> <br\/> <br\/> <span class='basic_pink'> V\u1eady $\\widehat{OBA} = 55^{o} \\\\ \\widehat{ABC} = 35^{o} \\\\ \\widehat{CDO} = 55^{o} $<\/span><\/span> <br\/>"}]}],"id_ques":1637},{"time":24,"part":[{"time":3,"title":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u","title_trans":" Cho tam gi\u00e1c ABC, M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia MA l\u1ea5y \u0111i\u1ec3m E sao cho ME = MA. <br\/> H\u00e3y s\u1eafp x\u1ebfp c\u00e1c \u00fd ch\u1ee9ng minh AB \/\/ CE. ","temp":"sequence","correct":[[[2],[3],[5],[4],[1]]],"list":[{"point":10,"image":"https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai11/lv3/img\/H7B11-11.png","left":["MB = MC (gt), $\\widehat{AMB} = \\widehat{EMC} $ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh), MA = ME (gi\u1ea3 thi\u1ebft)","Do \u0111\u00f3 $\\triangle AMB = \\triangle EMC (c.g.c) $","$\\widehat{MAB} = \\widehat{MEC} \\Rightarrow AB \/\/ CE $","$\\triangle AMB = \\triangle EMC \\Rightarrow \\widehat{MAB} = \\widehat{MEC}$ ","$\\triangle AMB $ v\u00e0 $\\triangle EMC $ c\u00f3: "],"top":55,"explain":" <span class='basic_left'>X\u00e9t $\\triangle AMB $ v\u00e0 $\\triangle EMC $ c\u00f3: <br\/> + $MB = MC$ (gi\u1ea3 thi\u1ebft)<br\/> + $\\widehat{AMB} = \\widehat{EMC} $ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh) <br\/> + $ MA = ME$ (gi\u1ea3 thi\u1ebft) <br\/> Do \u0111\u00f3 $\\triangle AMB = \\triangle EMC (c.g.c) $ <br\/>$ \\Rightarrow \\widehat{MAB} = \\widehat{MEC}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> M\u00e0 hai g\u00f3c $MAB$ v\u00e0 g\u00f3c $MEC$ \u1edf v\u1ecb tr\u00ed so le trong v\u1edbi nhau <br\/>$\\Rightarrow AB \/\/ CE $ <\/span> "}]}],"id_ques":1638},{"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":[[["35"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Cho $\\triangle ABC $ c\u00f3 $\\widehat{A} = 90^{o} $. Tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{B} $ c\u1eaft c\u1ea1nh AC t\u1ea1i \u0111i\u1ec3m D. Tr\u00ean c\u1ea1nh BC l\u1ea5y \u0111i\u1ec3m H sao cho $BH = BA.$ <br\/> Bi\u1ebft $\\widehat{ADH} = 110^{o} $. T\u00ednh s\u1ed1 \u0111o g\u00f3c $\\widehat{ABD} $ <br\/> <b>\u0110\u00e1p \u00e1n:<\/b> $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o}$ <\/span>","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai11/lv3/img\/H7B11-57.png' \/><\/center> <span class='basic_left'>X\u00e9t tam gi\u00e1c $\\triangle{BAD} $ v\u00e0 $\\triangle{BHD} $ c\u00f3: <br\/> + $BH = BA$ (gi\u1ea3 thi\u1ebft) <br\/> + $\\widehat{B_{1}} = \\widehat{B_{2}} $ (BD l\u00e0 tia ph\u00e2n gi\u00e1c) <br\/> + c\u1ea1nh BD chung <br\/> $\\Rightarrow \\triangle BAD = \\triangle BHD$ (c.g.c) <br\/> Suy ra $\\widehat{BDA} = \\widehat{BDH} $ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> M\u1eb7t kh\u00e1c: $\\widehat{ADH} = \\widehat{BDA} + \\widehat{BDH} = 110^{o} \\\\ \\Rightarrow \\widehat{BDA} = \\dfrac{110^{o}}{2} = 55^{o} $ <br\/> Tam gi\u00e1c $BAD$ vu\u00f4ng t\u1ea1i A n\u00ean: <br\/> $\\widehat{BAD} + \\widehat{ABD} = 90^{o}$ <br\/> $ \\Rightarrow \\widehat{ABD} = 90^{o} - \\widehat{BDA}$ <br\/> $ = 90^{o} - 55^{o} = 35^{o}$ <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0: 35<\/span> "}]}],"id_ques":1639},{"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 $ABC$. Tr\u00ean tia BA l\u1ea5y \u0111i\u1ec3m E sao cho A n\u1eb1m gi\u1eefa B v\u00e0 E, $AE = AC$. <br\/> Tr\u00ean tia $CA$ l\u1ea5y \u0111i\u1ec3m F sao cho A n\u1eb1m gi\u1eefa C v\u00e0 F, $AF = AB$. <br\/> K\u1ebb \u0111\u01b0\u1eddng trung tuy\u1ebfn $AM$ \u1ee9ng v\u1edbi c\u1ea1nh BC c\u1ee7a tam gi\u00e1c $ABC$. K\u1ebb \u0111\u01b0\u1eddng trung tuy\u1ebfn $AN$ \u1ee9ng v\u1edbi c\u1ea1nh $EF$ c\u1ee7a tam gi\u00e1c $AEF$. <br\/> S\u1ed1 c\u1eb7p tam gi\u00e1c b\u1eb1ng nhau l\u00e0: <\/span> ","select":[" A. 1 "," B. 2"," C. 3","D. 4 "],"explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai11/lv3/img\/H7B11-36.png' \/><\/center> X\u00e9t $\\triangle ABC $ v\u00e0 $\\triangle AEF $ c\u00f3: <br\/> + $AB = AF$ <br\/> + $\\widehat{BAC} = \\widehat{FAE} $ (\u0111\u1ed1i \u0111\u1ec9nh) <br\/> + $ AC = AE$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow \\triangle ABC = \\triangle AEF $ (c.g.c) <br\/> Suy ra $\\widehat{C} = \\widehat{E}, \\widehat{B} = \\widehat{F} $ <br\/> C\u00f3 $BC = EF \\Rightarrow BM = FN, MC = NE$ <br\/> T\u1eeb \u0111\u00f3 ta suy ra \u0111\u01b0\u1ee3c $\\triangle ABM = \\triangle AFN (c.g.c) \\\\ \\triangle AMC = \\triangle ANE (c.g.c)$ <br\/> Nh\u01b0 v\u1eady c\u00f3 3 c\u1eb7p tam gi\u00e1c b\u1eb1ng nhau. <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 C. <\/span>","column":4}]}],"id_ques":1640}],"lesson":{"save":0,"level":3}}