{"segment":[{"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":"Cho h\u00ecnh v\u1ebd, bi\u1ebft $CB = CD, BF = DE$. <br\/> H\u00e3y gh\u00e9p c\u00e1c \u00fd ch\u1ee9ng minh $\\widehat{C_{1}} = \\widehat{C_{2}} $","audio":"","temp":"matching","correct":[["2","3","4","1"]],"list":[{"point":10,"image":"https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai13/lv3/img\/H7B13-44.png","left":["$CB = CD, BF = DE $","$\\widehat{B} = \\widehat{D} $","$\\triangle CBF = \\triangle CDE $","$\\widehat{C_{1}} = \\widehat{C_{2}} $"],"right":["hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng","gi\u1ea3 thi\u1ebft","do $\\triangle CBD $ c\u00e2n","theo tr\u01b0\u1eddng h\u1ee3p c\u1ea1nh-g\u00f3c-c\u1ea1nh"],"top":100,"explain":"<span class='basic_left'>Ta c\u00f3: <br\/> + $CB = CD$ <br\/> + $\\widehat{B} = \\widehat{D} $ (do $\\triangle CBD $ c\u00e2n t\u1ea1i C ) <br\/> + $BF = DE $ (gi\u1ea3 thi\u1ebft) <br\/> Do \u0111\u00f3 $\\triangle CBF = \\triangle CDE $ (c.g.c) <br\/> Suy ra $\\widehat{C_{1}} = \\widehat{C_{2}} $ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng).<\/span> "}]}],"id_ques":1691},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["t","f","f","t","t"]],"list":[{"point":10,"image":"","ques":"","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["N\u1ebfu m\u1ed9t tam gi\u00e1c c\u00f3 hai c\u1ea1nh b\u1eb1ng nhau th\u00ec tam gi\u00e1c \u0111\u00f3 c\u00f3 hai g\u00f3c b\u1eb1ng nhau.","N\u1ebfu m\u1ed9t tam gi\u00e1c c\u00f3 m\u1ed9t g\u00f3c b\u1eb1ng $60^{o}$ th\u00ec tam gi\u00e1c \u0111\u00f3 l\u00e0 tam gi\u00e1c \u0111\u1ec1u.","N\u1ebfu m\u1ed9t tam gi\u00e1c c\u00f3 hai g\u00f3c b\u1eb1ng nhau v\u00e0 hai c\u1ea1nh b\u1eb1ng nhau th\u00ec \u0111\u00f3 l\u00e0 tam gi\u00e1c \u0111\u1ec1u. ","Tam gi\u00e1c vu\u00f4ng c\u00f3 hai g\u00f3c nh\u1ecdn b\u1eb1ng nhau l\u00e0 tam gi\u00e1c vu\u00f4ng c\u00e2n. ","Tam gi\u00e1c c\u00f3 hai g\u00f3c b\u1eb1ng $60^{o} $ l\u00e0 tam gi\u00e1c \u0111\u1ec1u."],"explain":["1 - \u0110\u00fang, theo \u0111\u1ecbnh ngh\u0129a v\u00e0 t\u00ednh ch\u1ea5t c\u1ee7a tam gi\u00e1c c\u00e2n.","<br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai13/lv3/img\/H7B13-42.png' \/><\/center> 2 - Sai. V\u00ed d\u1ee5 $\\triangle ABC$ c\u00f3 $\\widehat{B} = 60^{o}, BC > AB $ th\u00ec $\\triangle ABC $ kh\u00f4ng ph\u1ea3i l\u00e0 tam gi\u00e1c \u0111\u1ec1u. ","<br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai13/lv3/img\/H7B13-43.png' \/><\/center> 3 - Sai. V\u00ed d\u1ee5 $\\triangle ABC $ c\u00f3 $\\widehat{B} = \\widehat{C}, AB = AC $ nh\u01b0ng $\\triangle ABC $ kh\u00f4ng ph\u1ea3i tam gi\u00e1c \u0111\u1ec1u.","<br\/> 4 - \u0110\u00fang theo \u0111\u1ecbnh ngh\u0129a tam gi\u00e1c c\u00e2n.","<br\/> 5 - \u0110\u00fang, v\u00ec khi \u0111\u00f3 g\u00f3c c\u00f2n l\u1ea1i c\u0169ng b\u1eb1ng $60^{o} $ do \u0111\u00f3 n\u00f3 l\u00e0 tam gi\u00e1c \u0111\u1ec1u theo \u0111\u1ecbnh ngh\u0129a."]}]}],"id_ques":1692},{"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 g\u00f3c $\\widehat{xOy} = 120^{o} $, \u0111i\u1ec3m A thu\u1ed9c tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c \u0111\u00f3. <br\/> K\u1ebb $AB \\perp Ox (B \\in Ox) $, k\u1ebb $AC \\perp Oy (C \\in Oy)$. <br\/> Tam gi\u00e1c ABC l\u00e0: <\/span> ","select":["A. Tam gi\u00e1c c\u00e2n","B. Tam gi\u00e1c \u0111\u1ec1u","C. Tam gi\u00e1c vu\u00f4ng","D. Tam gi\u00e1c vu\u00f4ng c\u00e2n."],"explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai13/lv3/img\/H7B13-18.png' \/><\/center> <span class='basic_left'> Ta th\u1ea5y trong $\\Delta AOB$ v\u00e0 $\\Delta AOC$: <br\/> + $\\widehat{O_{1}} = \\widehat{O_{2}} $ (do OA l\u00e0 tia ph\u00e2n gi\u00e1c) <br\/> + $\\widehat{OCA} = \\widehat {OBA} = 90^{o}$ <br\/> $\\Rightarrow$ $\\widehat {CAO} = \\widehat {BAO}$ <br\/> X\u00e9t $ \\triangle AOB $ v\u00e0 $\\triangle AOC$ c\u00f3: <br\/> + $\\widehat{O_{1}} = \\widehat{O_{2}} $ (do OA l\u00e0 tia ph\u00e2n gi\u00e1c) <br\/> + C\u1ea1nh OA chung <br\/> + $\\widehat {CAO} = \\widehat {BAO}$ (ch\u1ee9ng minh tr\u00ean) <br\/> Do \u0111\u00f3 $\\triangle AOB = \\triangle AOC $ (g . c . g) <br\/> $\\Rightarrow AB = AC $ (1) <br\/> M\u1eb7t kh\u00e1c c\u00f3 $\\widehat{O_{1}} = \\widehat{O_{2}} = \\dfrac{xOy}{2} = \\dfrac{120^{o}}{2} = 60^{o} $ <br\/> n\u00ean $ \\widehat{A_{1}} = \\widehat{A_{2}} = 90^{o} - 60^{o} = 30^{o} $ <br\/> suy ra $\\widehat{BAC} = 60^{o} $ (2) <br\/> T\u1eeb (1) v\u00e0 (2), tam gi\u00e1c ABC c\u00e2n c\u00f3 $\\widehat{BAC} = 60^{o} $ n\u00ean l\u00e0 tam gi\u00e1c \u0111\u1ec1u.<br\/> <span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 B. <\/span>","column":2}]}],"id_ques":1693},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["t","f","t","t"]],"list":[{"point":10,"image":"","ques":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A (AB < AC)$. Tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c A c\u1eaft $BC$ t\u1ea1i D. Qua D k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi $BC$ c\u1eaft $AC$ t\u1ea1i E. Tr\u00ean $AB$ l\u1ea5y \u0111i\u1ec3m F sao cho $AF = AE$. Khi \u0111\u00f3: ","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["$\\widehat{B} = \\widehat{DEC} $","$\\triangle DBF $ c\u00e2n t\u1ea1i B","$DB = DF$","$DB = DE$"],"explain":["<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai13/lv3/img\/H7B13-21.png' \/><\/center> <span class='basic_left'> 1 - \u0110\u00fang. V\u00ec hai g\u00f3c c\u00f9ng ph\u1ee5 v\u1edbi g\u00f3c $\\widehat{C} $ <\/span>","<br\/> <span class='basic_left'> 2 - Sai. <br\/> X\u00e9t $\\Delta EAD$ v\u00e0 $\\Delta FAD$ c\u00f3: <br\/> + $AE=AF$ (gi\u1ea3 thi\u1ebft) <br\/> + $\\widehat{EAD}=\\widehat{FAD}$ (gi\u1ea3 thi\u1ebft) <br\/> + C\u1ea1nh AD chung <br\/> $\\Rightarrow \\triangle EAD = \\triangle FAD (c.g.c) $ <br\/> $\\Rightarrow \\widehat{E_{2}} = \\widehat{F_{2}} \\\\ \\Rightarrow \\widehat{E_{1}} = \\widehat{F_{1}} $ <br\/> M\u00e0 $\\widehat{E_{1}} = \\widehat{B}$ n\u00ean $\\widehat{F_{1}} = \\widehat{B} $ <br\/> Hay $\\triangle DBF $ c\u00e2n t\u1ea1i D.<\/span>","<br\/><span class='basic_left'> 3 - \u0110\u00fang. V\u00ec: $\\triangle DBF $ c\u00e2n t\u1ea1i D.<\/span> "," <br\/><span class='basic_left'> 4 - \u0110\u00fang. V\u00ec c\u00f9ng b\u1eb1ng DF.<\/span> "]}]}],"id_ques":1694},{"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 h\u00ecnh v\u1ebd: <br\/> S\u1ed1 \u0111o c\u1ee7a g\u00f3c $\\widehat{CDA} $ l\u00e0: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai13/lv3/img\/H7B13-26.png' \/><\/center> ","select":["A. $45^{o} $ ","B. $22,5^{o} $ ","C. $67,5^{o} $ ","D. $30^{o} $ "],"hint":"- T\u00ednh s\u1ed1 \u0111o c\u1ee7a g\u00f3c $\\widehat{ACD} $ <br\/> - D\u1ef1a v\u00e0o tam gi\u00e1c CAD c\u00e2n \u0111\u1ec3 suy ra s\u1ed1 \u0111o g\u00f3c $\\widehat{CDA} $","explain":"<span class='basic_left'> Ta c\u00f3 tam gi\u00e1c $ABC$ vu\u00f4ng c\u00e2n t\u1ea1i A, <br\/> $\\Rightarrow \\widehat{ACB} = \\widehat{ABC} = 45^{o} $ <br\/> C\u00f3 $\\widehat{ACB} + \\widehat{ACD} = 180^{o} $ (hai g\u00f3c k\u1ec1 b\u00f9) <br\/> $\\Rightarrow \\widehat{ACD} = 180^{o} - \\widehat{ACB} = 180^{o} - 45^{o} = 135^{o} $ <br\/> L\u1ea1i c\u00f3 tam gi\u00e1c $ACD$ c\u00e2n t\u1ea1i C n\u00ean: <br\/> $ \\widehat{CDA} = \\widehat{CAD} = \\dfrac{180^{o} -135^{o}}{2} = 22,5^{o} $ <br\/><br\/> <span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 B. <\/span>","column":4}]}],"id_ques":1695},{"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":[[["90"]]],"list":[{"point":10,"width":20,"content":"","type_input":"","ques":"<span class='basic_left'> Cho tam gi\u00e1c c\u00e2n $AOB$ $(OA = OB)$. <br\/> Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia OB l\u1ea5y \u0111i\u1ec3m C sao cho OB = OC <br\/> S\u1ed1 \u0111o g\u00f3c $\\widehat{BAC} $ 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/bai13/lv3/img\/H7B13-31.png' \/><\/center> <span class='basic_left'> $\\triangle AOB $ c\u00e2n t\u1ea1i O n\u00ean $\\widehat{A_{1}} = \\widehat{B} $ <br\/> $\\triangle AOC $ c\u00e2n t\u1ea1i O n\u00ean $\\widehat{A_{2}} = \\widehat{C} $ <br\/> Suy ra $\\widehat{A_{1}} + \\widehat{A_{2}} = \\widehat{B} + \\widehat{C} \\\\ \\Leftrightarrow \\widehat{BAC} = \\widehat{B} + \\widehat{C} \\hspace{0,2cm} (1) $ <br\/> Tam gi\u00e1c $ABC$ c\u00f3: $\\widehat{BAC} + \\widehat{B} + \\widehat{C} = 180^{o} $ (2) <br\/> T\u1eeb (1) v\u00e0 (2) suy ra $2\\widehat{BAC} = 180^{o} \\\\ \\Rightarrow \\widehat{BAC} = 90^{o} $<br\/><br\/> <span class='basic_pink'>V\u1eady $\\widehat{BAC} = 90^{o} $<\/span>"}]}],"id_ques":1696},{"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'> Tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i A, \u0111i\u1ec3m M thu\u1ed9c c\u1ea1nh BC. K\u1ebb $MD \\perp AB (D \\in AB) $, k\u1ebb $ME \\perp AC (E \\in AC)$, k\u1ebb $BH \\perp AC (H \\in AC) $. Khi \u0111\u00f3: ","select":[" $MD + ME < BH$ "," $MD + ME = BH$ "," $MD + ME > BH$ "],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai13/lv3/img\/H7B13-32.png' \/><\/center> <span class='basic_left'>K\u1ebb $MK \\perp BH. $ Khi \u0111\u00f3 $ME \/\/ KH, MK \/\/ EH. $ <br\/> *X\u00e9t $\\triangle KMH $ v\u00e0 $\\triangle EHM $ c\u00f3: <br\/> + $\\widehat{H_{2}} = \\widehat{M_{2}} $ (so le trong)<br\/> + c\u1ea1nh MH chung<br\/> + $\\widehat{M_{1}} = \\widehat{H_{1}} $ (so le trong) <br\/>$\\Rightarrow \\triangle KMH = \\triangle EHM (c.g.c) $ <br\/> $\\Rightarrow ME = KH (1) $ <br\/> * Ta c\u00f3 $\\triangle ABC $ c\u00e2n t\u1ea1i A n\u00ean $\\widehat{B} = \\widehat{C}. $ <br\/> L\u1ea1i c\u00f3 $MK \/\/ AC$ $\\Rightarrow \\widehat{BMK} = \\widehat{C} $ (\u0111\u1ed3ng v\u1ecb) <br\/> $\\Rightarrow \\widehat{B} = \\widehat{BMK} $ <br\/> X\u00e9t tam gi\u00e1c vu\u00f4ng $MBD$ v\u00e0 $BMK$ c\u00f3: <br\/> + c\u1ea1nh huy\u1ec1n BM chung<br\/> + $ \\widehat{B} = \\widehat{BMK} $ <br\/> $\\Rightarrow \\triangle MBD = \\triangle BMK $ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow MD = BK $ (2) <br\/> * T\u1eeb (1) v\u00e0 (2) ta c\u00f3 $MD + ME = BK + KH = BH $ <br\/> Hay $MD + ME = BH $<\/span> ","column":3}]}],"id_ques":1697},{"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":[[["120"]]],"list":[{"point":10,"width":20,"content":"","type_input":"","ques":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$.<br\/> \u1ede ph\u00eda ngo\u00e0i tam gi\u00e1c $ABC$ v\u1ebd c\u00e1c tam gi\u00e1c \u0111\u1ec1u $ABD$ v\u00e0 $ACE$. <br\/>G\u1ecdi I l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $DC$ v\u00e0 $BE$. <br\/> 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/bai13/lv3/img\/H7B13-33.png' \/><\/center><span class='basic_left'>Ta c\u00f3 $\\triangle ADB $ \u0111\u1ec1u n\u00ean $\\widehat{DAB} = 60^{o} $, $\\triangle ACE $ \u0111\u1ec1u n\u00ean $\\widehat{CAE} = 60^{o}$ <br\/> $\\Rightarrow \\widehat{DAC} = \\widehat{BAE} (= 60^{o} + \\widehat{BAC}) $ <br\/> X\u00e9t $\\triangle ADC $ v\u00e0 $\\triangle ABE $ c\u00f3: <br\/> + $AD = AB$ (gi\u1ea3 thi\u1ebft)<br\/> + $ \\widehat{DAC} = \\widehat{BAE}$<br\/> + $AC = AE$ <br\/> Do \u0111\u00f3 $\\triangle ADC = \\triangle ABE (c.g.c) $ <br\/> $\\Rightarrow \\widehat{D_{1}} = \\widehat{B_{1}} $ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> G\u1ecdi K l\u00e0 giao \u0111i\u1ec3m c\u1ee7a AB v\u00e0 CD. <br\/> Tam gi\u00e1c $KAD$ v\u00e0 $KIB$ c\u00f3: <br\/> + $\\widehat{D_{1}} = \\widehat{B_{1}} $ (ch\u1ee9ng minh tr\u00ean)<br\/> + $\\widehat{K_{1}} = \\widehat{K_{2}} $ (\u0111\u1ed1i \u0111\u1ec9nh) <br\/> $\\Rightarrow \\widehat{KAD} = \\widehat{KIB} $ <br\/> M\u00e0 $\\widehat{KAD} = 60^{o} \\Rightarrow \\widehat{KIB} = 60^{o} $ <br\/> Ta c\u00f3 $\\widehat{KIB} + \\widehat{BIC} = 180^{o} $ (k\u1ec1 b\u00f9) <br\/> $\\Rightarrow \\widehat{BIC} = 180^{o} - 60^{o} = 120^{o} $ <br\/><br\/> <span class='basic_pink'>V\u1eady $\\widehat{BIC} = 120^{o} $<\/span>"}]}],"id_ques":1698},{"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":[[["115"]]],"list":[{"point":10,"width":20,"content":"","type_input":"","ques":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i A c\u00f3 $\\widehat{BAC} = 50^{o} $. <br\/> Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia BC l\u1ea5y \u0111i\u1ec3m D, tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia CB l\u1ea5y \u0111i\u1ec3m E sao cho $BD = BA, CE = CA.$ <br\/> S\u1ed1 \u0111o g\u00f3c $\\widehat{DAE} $ 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/bai13/lv3/img\/H7B13-40.png' \/><\/center> <span class='basic_left'> Tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i A c\u00f3 $\\widehat{A} = 50^{o} $ n\u00ean $\\widehat{B_{1}} = \\widehat{C_{1}} = \\dfrac{180^{o}-50^{o}}{2} = 65^{o} $ <br\/> Ta c\u00f3 $BD = BA$ (gi\u1ea3 thi\u1ebft) n\u00ean $\\triangle ABD $ c\u00e2n t\u1ea1i B $\\Rightarrow \\widehat{A_{1}} = \\widehat{D} $ <br\/> M\u00e0 $\\widehat{A_1}+\\widehat{D}=\\widehat{B_1}$ (t\u00ednh ch\u1ea5t g\u00f3c ngo\u00e0i tam gi\u00e1c) <br\/> $\\Rightarrow \\widehat{A_{1}} = \\widehat{D} = \\dfrac{1}{2}\\widehat{B_{1}} $ <br\/> T\u01b0\u01a1ng t\u1ef1, ta c\u00f3 $\\widehat{A_{2}} = \\widehat{E} = \\dfrac{1}{2}\\widehat{C_{1}} $ <br\/> T\u1eeb \u0111\u00f3 ta c\u00f3: $\\widehat{D} + \\widehat{E} = \\dfrac{1}{2}\\widehat{B_{1}} + \\dfrac{1}{2}\\widehat{C_{1}} = \\widehat{B_{1}} = 65^{o} $ <br\/> Tam gi\u00e1c $AED$ c\u00f3 $\\widehat{DAE} + \\widehat{D} + \\widehat{E} = 180^{o} \\\\ \\Rightarrow \\widehat{DAE} = 180^{o} - (\\widehat{D} + \\widehat{E}) \\\\ = 180^{o} - 65^{o} \\\\ = 115^{o} $<\/span> <br\/><br\/> <span class='basic_pink'>V\u1eady $\\widehat{DAE} = 115^{o} $<\/span>"}]}],"id_ques":1699},{"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":[[["50"]]],"list":[{"point":10,"width":20,"content":"","type_input":"","ques":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ nh\u01b0 h\u00ecnh v\u1ebd, c\u00f3 $\\widehat{A} = 80^{o} $. <br\/> Bi\u1ebft r\u1eb1ng $BO = BM, CO = CN $ <br\/> S\u1ed1 \u0111o g\u00f3c $\\widehat{MON} $ l\u00e0: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o}$ <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai13/lv3/img\/H7B13-41.png' \/><\/center> <\/span>","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai13/lv3/img\/H7B13-41a.png' \/><\/center> Tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{A} = 80^{o} $ n\u00ean $\\widehat{B} + \\widehat{C} = 100^{o} $ <br\/> Ta c\u00f3 $BO = BM$ (gi\u1ea3 thi\u1ebft) n\u00ean $\\triangle BOM $ c\u00e2n t\u1ea1i B $\\Rightarrow \\widehat{O_{1}} = \\widehat{M_{1}} $ <br\/> $\\Rightarrow \\widehat{O_{1}} = \\dfrac{180^{o}-\\widehat{B}}{2} $ <br\/> T\u01b0\u01a1ng t\u1ef1 c\u00f3: $ \\widehat{O_{2}} = \\dfrac{180^{o}-\\widehat{C}}{2} $ <br\/> Tam gi\u00e1c $MON$ c\u00f3 $\\widehat{O_{1}} + \\widehat{MON} + \\widehat{O_{2}} = 180^{o} $ <br\/> $ \\Rightarrow \\widehat{MON} $$ = 180^{o} - (\\widehat{O_{1}} + \\widehat{O_{2}}) \\\\ = 180^{o} - \\left( \\dfrac{180^{o}-\\widehat{B}}{2} + \\dfrac{180^{o}-\\widehat{C}}{2} \\right) \\\\ = 180^{o} - \\dfrac{360^{o}-\\widehat{B}-\\widehat{C}}{2} \\\\ = \\dfrac{\\widehat{B}+\\widehat{C}}{2} \\\\ = \\dfrac{100^{o}}{2} \\\\ = 50^{o} $<br\/> <span class='basic_pink'>V\u1eady $\\widehat{MON} = 50^{o} $<\/span>"}]}],"id_ques":1700}],"lesson":{"save":0,"level":3}}