{"common":{"save":0,"post_id":"1002","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"1751","post_id":"1002","mon_id":"0","chapter_id":"0","question":"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","options":{"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 $\\triangle ABC $. C\u00e1c tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c B v\u00e0 C c\u1eaft nhau \u1edf I. <br\/> K\u1ebb $ID \\perp AB (D \\in AB), IE \\perp AC (E \\in AC), IH \\perp AB (H \\in AB) $. Khi \u0111\u00f3: ","audio":"","temp":"matching","correct":[["4","3","1","2"]],"list":[{"point":10,"image":"img\/H7B15-18.png","left":["$ID = IH$ ","$IE = IH$ ","$ID = IE$","$AD = AE$ "],"right":["V\u00ec c\u00f9ng b\u1eb1ng \u0111\u1ed9 d\u00e0i $IH$","V\u00ec $\\triangle IAD = \\triangle IAE $ (c\u1ea1nh huy\u1ec1n - c\u1ea1nh g\u00f3c vu\u00f4ng).","V\u00ec $\\triangle CIE = \\triangle CIH $ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn).","V\u00ec $\\triangle BID = \\triangle BIH $ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn)."],"top":100,"explain":"<span class='basic_left'> Ta d\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c: $\\triangle BID = \\triangle BIH $ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow ID = IH$ (1) <br\/> T\u01b0\u01a1ng t\u1ef1: $\\triangle CIE = \\triangle CIH $ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow IE = IH$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) suy ra $ID = IE (= IH)$ <br\/> Do \u0111\u00f3: <br\/> + $ID=IE$ (ch\u1ee9ng minh tr\u00ean) <br\/> + $AD$ l\u00e0 c\u1ea1nh chung <br\/> + $\\widehat{ADI}=\\widehat{AEI}=90^o$ <br\/> $\\Rightarrow \\triangle IAD = \\triangle IAE $ (c\u1ea1nh huy\u1ec1n - c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow AD = AE $ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng)"}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:23:03"},{"id":"1752","post_id":"1002","mon_id":"0","chapter_id":"0","question":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","options":{"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":"<b>T\u00ecm c\u00e2u tr\u1ea3 l\u1eddi sai <\/b> <br\/> Cho $\\triangle ABC $ vu\u00f4ng t\u1ea1i A c\u00f3 $AC = \\dfrac{1}{2}BC $, M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh BC. Ta c\u00f3: ","select":["A. $MA = MB = MC $ ","B. $\\widehat{ABC} = 30^{o} $","C. $\\triangle AMC $ \u0111\u1ec1u ","D. $\\triangle AMB$ \u0111\u1ec1u "],"hint":"Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia AC l\u1ea5y \u0111i\u1ec3m D sao cho AD = AC.","explain":" <center><img src='img\/H7B15-25.png' \/><\/center> <span class='basic_left'>* Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $AC$ l\u1ea5y \u0111i\u1ec3m D sao cho $AD = AC$. <br\/> Khi \u0111\u00f3, ta d\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c: $\\triangle BAD = \\triangle BAC (c.g.c) $ <br\/> $\\Rightarrow BD = BC $ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> V\u00ec $AC = \\dfrac{1}{2}BC \\Rightarrow BC = 2AC $ <br\/> V\u00ec $AD = AC \\Rightarrow CD = 2AC $ <br\/> Do \u0111\u00f3 $BC = CD$ (2)<br\/> T\u1eeb (1) v\u00e0 (2) $BD = BC = CD$ <br\/> $\\Rightarrow \\triangle BCD $ \u0111\u1ec1u $\\Rightarrow \\widehat{MCA} = 60^{o} $ n\u00ean $\\widehat{BCA} = 60^{o} $ <br\/> Ta c\u00f3: $AC = MC \\left(= \\dfrac{BC}{2} \\right)$ v\u00e0 $\\widehat{MCA} = 60^{o} $ <br\/> $\\Rightarrow AMC $ \u0111\u1ec1u $\\Rightarrow $ C\u00e2u C \u0111\u00fang. <br\/> * X\u00e9t $\\triangle ABC $ vu\u00f4ng t\u1ea1i A ta c\u00f3: <br\/> $\\widehat{ABC} + \\widehat{BCA} = 90^{o} $<br\/> $\\Rightarrow \\widehat{ABC} = 90^{o} - \\widehat{BCA} = 30^{o} \\Rightarrow$ C\u00e2u B \u0111\u00fang. <br\/> * V\u00ec M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC n\u00ean $MB = MC = \\dfrac{BC}{2}$ <br\/> V\u00ec $\\triangle AMC $ \u0111\u1ec1u n\u00ean MC = MA <br\/> $\\Rightarrow MA = MB = MC \\Rightarrow $ C\u00e2u A \u0111\u00fang. <br\/> Nh\u01b0 v\u1eady: C\u1ea3 A, B, C \u0111\u1ec1u \u0111\u00fang. <\/span> <br\/><br\/><span class='basic_pink'>C\u00e2u tr\u1ea3 l\u1eddi sai l\u00e0 D. <\/span>","column":2}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:23:03"},{"id":"1753","post_id":"1002","mon_id":"0","chapter_id":"0","question":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u","options":{"time":24,"part":[{"time":3,"title":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u","title_trans":"H\u00e3y s\u1eafp x\u1ebfp c\u00e1c \u00fd ch\u1ee9ng minh $MN$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c M. ","temp":"sequence","correct":[[[4],[3],[2],[1],[5]]],"list":[{"point":10,"image":"img\/H7B15-15.png","left":["Suy ra $\\widehat{M_{1}} = \\widehat{M_{2}} $","Do \u0111\u00f3 $\\triangle DMN = \\triangle CMN $ (c\u1ea1nh huy\u1ec1n - c\u1ea1nh g\u00f3c vu\u00f4ng)","$DN = CN$ (gi\u1ea3 thi\u1ebft), c\u1ea1nh $MN$ chung","X\u00e9t hai tam gi\u00e1c vu\u00f4ng $DMN$ v\u00e0 $CMN$, ta c\u00f3: ","V\u1eady $MN$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c M. "],"top":55,"explain":"<span class='basic_left'> X\u00e9t hai tam gi\u00e1c vu\u00f4ng $DMN$ v\u00e0 $CMN$, ta c\u00f3: <br\/>+ $ DN = CN $(gi\u1ea3 thi\u1ebft)<br\/> + c\u1ea1nh $MN$ chung <br\/> Do \u0111\u00f3 $\\triangle DMN = \\triangle CMN $ (c\u1ea1nh huy\u1ec1n - c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/>Suy ra $\\widehat{M_{1}} = \\widehat{M_{2}} $ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> V\u1eady $MN$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c M. "}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:23:03"}]}