{"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 $\\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) $. Khi \u0111\u00f3: ","audio":"","temp":"matching","correct":[["4","3","1","2"]],"list":[{"point":10,"image":"https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai15/lv3/img\/H7B15-18.png","left":["$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)"}]}],"id_ques":1751},{"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='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai15/lv3/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}]}],"id_ques":1752},{"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":"https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai15/lv3/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. "}]}],"id_ques":1753},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","temp":"true_false","correct":[["t","t","t","f"]],"list":[{"point":10,"ques":" <span class='basic_left'>Cho tam gi\u00e1c $ABC$ c\u00f3 $AB < AC$. Tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c A c\u1eaft \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a BC t\u1ea1i I. Qua I k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi hai c\u1ea1nh c\u1ee7a g\u00f3c A, c\u1eaft c\u00e1c tia $AB, AC$ theo th\u1ee9 t\u1ef1 t\u1ea1i H v\u00e0 K. Khi \u0111\u00f3: ","image":"","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["$AH = AK$","$BH = CK $","$AK = \\dfrac{AC + AB}{2} $","$CK = \\dfrac{AC+AB}{2} $"],"explain":["<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai15/lv3/img\/H7B15-43.png' \/><\/center> <span class='basic_left'> 1 - \u0110\u00fang. V\u00ec ta ch\u1ee9ng minh \u0111\u01b0\u1ee3c ngay: $\\triangle AHI = \\triangle AKI $ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow AH = AK $","<span class='basic_left'>2 - \u0110\u00fang. G\u1ecdi M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$. <br\/> $\\triangle BMI = \\triangle CMI (c.g.c) \\Rightarrow IB = IC $ <br\/> $\\triangle AHI = \\triangle AKI \\Rightarrow IH = IK$ <br\/> $\\triangle IHB = \\triangle IKC $ (c\u1ea1nh huy\u1ec1n - c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/>$\\Rightarrow BH = CK$ ","<span class='basic_left'>3 - \u0110\u00fang, v\u00ec $AC = AK + KC (1) \\\\ AB = AH - BH (2)$ <br\/> T\u1eeb (1) v\u00e0 (2) suy ra: <br\/> $AC + AB = (AK + AH) + (KC - BH)$ <br\/> Do $AH = AK, BH = CK$ n\u00ean $AC + AB = 2AK$ <br\/> $\\Rightarrow AK = \\dfrac{AC + AB}{2} $ ","<span class='basic_left'>4 - Sai. V\u00ec: T\u1eeb (1) v\u00e0 (2) ta suy ra: <br\/> $AC - AB = (AK - AH) + (KC + BH) \\\\ \\Rightarrow CK = \\dfrac{AC-AB}{2} $ "]}]}],"id_ques":1754},{"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 $ABC$ vu\u00f4ng c\u00e2n t\u1ea1i $A$ c\u00f3 $O$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh $BC$. <br\/> Qua $O$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng d b\u1ea5t k\u1ef3 kh\u00f4ng qua \u0111\u1ec9nh c\u1ee7a tam gi\u00e1c. K\u1ebb $BI, AH, CK$ vu\u00f4ng g\u00f3c v\u1edbi d. <br\/> T\u00ednh $BI^{2} + CK^{2} + 2.AH^{2} $, bi\u1ebft r\u1eb1ng $BC = 10cm.$ <br\/> \u0110\u00e1p \u00e1n: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} (cm)$ ","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai15/lv3/img\/H7B15-44.png' \/><\/center> <span class='basic_left'> Ta c\u00f3: $AB=AC; BO=CO; AO$ chung <br\/> $\\Rightarrow \\triangle ABO = \\triangle ACO (c.c.c) \\\\ \\Rightarrow \\widehat{AOB} = \\widehat{AOC} = 90^{o} $ <br\/> M\u1eb7t kh\u00e1c: $\\Delta ABC$ vu\u00f4ng c\u00e2n t\u1ea1i A n\u00ean $\\widehat{ABC}= 45^{o} $<br\/> $\\Rightarrow \\Delta ABO$ vu\u00f4ng c\u00e2n t\u1ea1i O<br\/> Do \u0111\u00f3 $BO = AO$. <br\/> V\u00ec: + $BO=CO$ (gi\u1ea3 thi\u1ebft) <br\/> + $\\widehat{BIO}=\\widehat{CKO}=90^o$ <br\/> + $\\widehat{BOI}=\\widehat{COK}$ (\u0111\u1ed1i \u0111\u1ec9nh) <br\/> $\\Rightarrow \\triangle BIO = \\triangle CKO $ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) suy ra $BI = CK$ <br\/> Ta c\u00f3: $\\widehat{BOI} + \\widehat{HOA} = 90^{o}; \\widehat{HAO} + \\widehat{HOA} = 90^{o} $ n\u00ean $\\widehat{BOI} = \\widehat{HAO} $ (1) <br\/> M\u00e0 $\\Delta ABC$ vu\u00f4ng c\u00e2n t\u1ea1i C v\u00e0 $BO=CO$ n\u00ean $BO=OA$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) suy ra $\\triangle BIO = \\triangle OHA $ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow BI = OH$ <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Py-ta-go, ta c\u00f3: <br\/> $BI^{2} + CK^{2} + 2.AH^{2} = BI^{2} + BI^{2} + 2.AH^{2} \\\\ = 2(OH^{2}+AH^{2}) = 2.AO^{2} = 2.25 = 50 (cm) $ <br\/> V\u1eady $BI^{2} + CK^{2} + 2.AH^{2} = 50(cm)$ <br\/> <span class='basic_pink'>\u0110\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0: 50<\/span> "}]}],"id_ques":1755},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","temp":"true_false","correct":[["t","t","t","t"]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i A c\u00f3 $\\widehat{A} < 90^{o} $. K\u1ebb $BD \\perp AC $ t\u1ea1i D, k\u1ebb $CE \\perp AB $ t\u1ea1i E. <br\/> G\u1ecdi K l\u00e0 giao \u0111i\u1ec3m c\u1ee7a c\u1ee7a $BD$ v\u00e0 $CE, I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$. Khi \u0111\u00f3: ","image":"","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["$\\triangle BCE = \\triangle CBD $ ","$\\triangle BEK = \\triangle CDK $ ","$AK$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BAC} $","Ba \u0111i\u1ec3m $A, K, I$ th\u1eb3ng h\u00e0ng"],"explain":["<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai15/lv3/img\/H7B15-45.png' \/><\/center> <span class='basic_left'> 1 - \u0110\u00fang. <br\/> $\\triangle BCE = \\triangle CBD $ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn), v\u00ec c\u00f3: <br\/> + $BC$ chung <br\/> + $\\widehat{CBE}=\\widehat{BCD}$ (gi\u1ea3 thi\u1ebft)<\/span>","<span class='basic_left'>2 - \u0110\u00fang. <br\/> V\u00ec $\\triangle BCE = \\triangle CBD$ (ch\u1ee9ng minh tr\u00ean) <br\/> $\\Rightarrow BE = CD $ (1) <br\/> M\u00e0 $\\widehat{EKB} = \\widehat{DKC}$ (\u0111\u1ed1i \u0111\u1ec9nh) <br\/>$ \\Rightarrow \\widehat{EBK} = \\widehat{DCK} $ (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\triangle BEK = \\triangle CDK $ (c\u1ea1nh g\u00f3c vu\u00f4ng - g\u00f3c nh\u1ecdn k\u1ec1) <\/span>","<span class='basic_left'>3 - \u0110\u00fang, v\u00ec: + $AK$ chung <br\/> + $AB=AC$ (gi\u1ea3 thi\u1ebft) <br\/> + $BK=KC$ (v\u00ec $\\triangle BEK = \\triangle CDK $) <br\/> $\\Rightarrow \\triangle ABK = \\triangle ACK (c.c.c) $ <br\/> $\\Rightarrow \\widehat{BAK}=\\widehat{CAK}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow $ $AK$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BAC} $ <\/span>","<span class='basic_left'>4 - \u0110\u00fang. <br\/> $AK$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BAC} $ (ch\u1ee9ng minh tr\u00ean) (3) <br\/> Theo b\u00e0i: $\\Delta ABC$ c\u00e2n t\u1ea1i A, m\u00e0 $IB=IC$ <br\/> Do \u0111\u00f3: $AI$ v\u1eeba l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn v\u1eeba l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $BAC$ <br\/> V\u1eady $AI$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BAC} $ (4) <br\/> T\u1eeb (3) v\u00e0 (4) $\\Rightarrow A, K, I $ th\u1eb3ng h\u00e0ng <\/span>"]}]}],"id_ques":1756},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","temp":"true_false","correct":[["t","t","t","f"]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ vu\u00f4ng c\u00e2n t\u1ea1i $A, N$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$, l\u1ea5y \u0111i\u1ec3m I n\u1eb1m gi\u1eefa N v\u00e0 C. K\u1ebb $BE$ v\u00e0 $CH$ c\u00f9ng vu\u00f4ng g\u00f3c v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng $AI$ ($E$ v\u00e0 $H$ thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng $AI$). Khi \u0111\u00f3: ","image":"","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["$BE = AH$","$\\triangle NAE = \\triangle NCH $","$\\widehat{ENH} = \\widehat{ANC} = 90^{o} $","$\\triangle NEH $ vu\u00f4ng c\u00e2n t\u1ea1i H"],"explain":["<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai15/lv3/img\/H7B15-47.png' \/><\/center><span class='basic_left'> 1 - \u0110\u00fang. Ta c\u00f3: <br\/> $\\widehat{A_1}+\\widehat{BAE}=90^o$ <br\/> $\\widehat{B_1}+\\widehat{BAE}=90^o$ <br\/> $\\Rightarrow \\widehat{A_1}=\\widehat{B_1}$ (1) <br\/> M\u00e0 $AB=AC$ (gi\u1ea3 thi\u1ebft) (2)<br\/> T\u1eeb (1) v\u00e0 (2) $\\triangle ABE = \\triangle CAH $ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow BE = AH $ <\/span>","<span class='basic_left'>2 - \u0110\u00fang. <br\/> Ch\u1ee9ng minh ngay \u0111\u01b0\u1ee3c $\\triangle ANC $ v\u00e0 $\\Delta ANB$ vu\u00f4ng c\u00e2n t\u1ea1i N <br\/> $\\Rightarrow AN = NC $ <br\/> $\\Rightarrow \\widehat{A_3}=\\widehat{C_1}=45^o$ (3) <br\/> Ta c\u00f3: $\\widehat{BAE}=\\widehat{ACH}$ <br\/> $\\Rightarrow \\widehat{A_2}+\\widehat{A_3}=\\widehat{C_1}+\\widehat{C_2}$ (v\u00ec $\\triangle ABE = \\triangle CAH $) (4) <br\/> T\u1eeb (3) v\u00e0 (4) suy ra: $\\widehat{A_2}=\\widehat{C_2}$ <br\/> T\u1eeb \u0111\u00f3, ta c\u00f3: <br\/> + $AN=CN$ <br\/> + $\\widehat{A_2}=\\widehat{C_2}$ <br\/> + $AE=CH$ (v\u00ec $\\triangle ABE = \\triangle CAH $) <br\/> $\\triangle NAE = \\triangle NCH (c.g.c) $<\/span>","<span class='basic_left'>3 - \u0110\u00fang. T\u1eeb $\\triangle NAE = \\triangle NCH$ (ch\u1ee9ng minh tr\u00ean) <br\/> $ \\Rightarrow NE = NH, \\widehat{ANE} = \\widehat{CNH} $ <br\/> T\u1eeb $\\widehat{ANE} = \\widehat{CNH} \\Rightarrow \\widehat{ENH} = \\widehat{ANC} = 90^{o} $ <\/span>","<span class='basic_left'><br\/> 4 - Sai. V\u00ec \u1edf h\u00e0ng th\u1ee9 3 ta \u0111\u00e3 ch\u1ee9ng minh \u0111\u01b0\u1ee3c: $NE=NH$ v\u00e0 $\\widehat{ENH}=90^o$ <br\/> $\\Rightarrow \\triangle ENH $ vu\u00f4ng c\u00e2n t\u1ea1i N. <\/span>"]}]}],"id_ques":1757},{"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 c\u00e2n $ABC (AB = AC). D$ l\u00e0 m\u1ed9t \u0111i\u1ec3m tr\u00ean $AB, E$ l\u00e0 m\u1ed9t \u0111i\u1ec3m tr\u00ean $AC$ sao cho $AD = AE$. T\u1eeb D v\u00e0 E h\u1ea1 c\u00e1c \u0111\u01b0\u1eddng $DM$ v\u00e0 $EN$ c\u00f9ng vu\u00f4ng g\u00f3c v\u1edbi \u0111\u00e1y $BC$. Khi \u0111\u00f3: ","select":["$BM < CN$","$BM = CN$","$BM > CN$"],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai15/lv3/img\/H7B15-48.png' \/><\/center> <span class='basic_left'> Tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i A n\u00ean $AB = AC$, v\u00e0 $AD = AE$ (gi\u1ea3 thi\u1ebft) <br\/> Do \u0111\u00f3 $ AB - AD = AC - AE \\\\ \\Rightarrow DB = EC $ <br\/> X\u00e9t tam gi\u00e1c vu\u00f4ng $DBM$ v\u00e0 $ECN$ c\u00f3: <br\/> + $DB = EC$ <br\/> + $\\widehat{B} = \\widehat{C} $ (do $\\triangle ABC$ c\u00e2n ) <br\/> $\\Rightarrow \\triangle DBM = \\triangle ECN $ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) <br\/> T\u1eeb \u0111\u00f3 suy ra $BM = CN$ ","column":3}]}],"id_ques":1758},{"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":"<b>Ph\u00e1t bi\u1ec3u sau ''\u0110\u00fang'' hay ''Sai'' <\/b> <br\/> ''Trong m\u1ed9t tam gi\u00e1c c\u00e2n, \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c \u1edf \u0111\u1ec9nh l\u00e0 \u0111\u01b0\u1eddng cao \u1ee9ng v\u1edbi \u0111\u00e1y c\u1ee7a tam gi\u00e1c \u0111\u00f3'' ","select":["\u0110\u00fang","Sai"],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai15/lv3/img\/H7B15-50.png' \/><\/center><span class='basic_left'> Gi\u1ea3 s\u1eed $AD$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c \u1edf \u0111\u1ec9nh A c\u1ee7a tam gi\u00e1c c\u00e2n $ABC$. <br\/> X\u00e9t $\\triangle ABD $ v\u00e0 $\\triangle ACD $ c\u00f3: <br\/> + $\\widehat{BAD} = \\widehat{CAD} $ (AD l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c A) <br\/> + $AB = AC$ ($\\triangle ABC $ c\u00e2n t\u1ea1i A) <br\/> + $\\widehat{B} = \\widehat{C} (\\triangle ABC)$ c\u00e2n t\u1ea1i A <br\/> Do \u0111\u00f3 $\\triangle ABD = \\triangle ACD (g.c.g) $ <br\/> $\\Rightarrow \\widehat{ADB} = \\widehat{ADC} $ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> Ba \u0111i\u1ec3m $B, D, C$ th\u1eb3ng h\u00e0ng n\u00ean $\\widehat{ADB} = \\widehat{ADC} = 90^{o}$ <br\/> hay $AD \\perp BC $ <br\/> <span class='basic_pink'> Ph\u00e1t bi\u1ec3u tr\u00ean l\u00e0 \u0110\u00fang. ","column":2}]}],"id_ques":1759},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","temp":"true_false","correct":[["t","t","t","t"]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho $\\Delta ABC$ c\u00e2n t\u1ea1i A. K\u1ebb $CH \\bot AB\\,(H\\in AB)$. K\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng d qua C vu\u00f4ng g\u00f3c v\u1edbi $AC$. K\u1ebb $BK\\bot d \\,(K\\in d)$. Khi \u0111\u00f3: ","image":"","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["$BK\/\/AC$ ","$\\widehat{ABC}=\\widehat{CBK}$ ","$\\Delta BHC=\\Delta BKC $","$CK=CH$"],"explain":["<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai15/lv3/img\/H7B15-57.jpg' \/><\/center> <span class='basic_left'> 1 - \u0110\u00fang. <br\/> Theo b\u00e0i: $BK \\bot CK$ <br\/> $AC\\bot CK$ <br\/> $\\Rightarrow BK \/\/AC$<\/span>","<span class='basic_left'>2 - \u0110\u00fang. <br\/> Do $BK\/\/AC$ (ch\u1ee9ng minh tr\u00ean) <br\/> $\\Rightarrow \\widehat{CBK}=\\widehat{ACB}$ (hai g\u00f3c so le trong) <br\/> M\u00e0 $\\widehat{ABC}=\\widehat{ACB}$ (do $\\Delta ABC$ c\u00e2n t\u1ea1i A) <br\/> $\\Rightarrow \\widehat{ABC}=\\widehat{CBK}$ <\/span>","<span class='basic_left'>3 - \u0110\u00fang <br\/> X\u00e9t hai tam gi\u00e1c vu\u00f4ng $BHC$ v\u00e0 $BKC$: <br\/> + $BC$ chung <br\/> + $\\widehat{HBC}=\\widehat{KBC}$ (ch\u1ee9ng minh tr\u00ean) <br\/> $\\Rightarrow \\triangle BHC = \\triangle BKC$ ( c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) <\/span>","<span class='basic_left'>4 - \u0110\u00fang. <br\/> Ta \u0111\u00e3 ch\u1ee9ng minh \u0111\u01b0\u1ee3c: $\\triangle BHC = \\triangle BKC$ <br\/> $\\Rightarrow CK=CH$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <\/span>"]}]}],"id_ques":1760}],"lesson":{"save":0,"level":3}}