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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 7: Cho ABC (AB

Toán Lớp 7: Cho ABC (AB

Comments ( 1 )

  1. a) Xét $\triangle$AMB và $\triangle$AMD có:
    AB = AD
    $\widehat{BAM}$ = $\widehat{DAM}$ (vì AM là phân giác)
    AM là cạnh chung
    Vậy $\triangle$AMB = $\triangle$AMD (c.g.c)
    => MB = MD ; $\widehat{ABM}$ = $\widehat{ADM}$
    b) Xét $\triangle$DAK và $\triangle$BAC có:
    $\widehat{ADK}$ = $\widehat{ABC}$ (câu a)
    AD = AB (gt)
    $\widehat{BAC}$ chung
    Vậy $\triangle$DAK = $\triangle$BAC (g.c.g)
    c) Vì $\triangle$DAK = $\triangle$BAC (cmt)
    -> AK = AC
    -> $\triangle$AKC cân tại A
    d) Vì AM là phân giác
    -> $\dfrac{MB}{MC}$ = $\dfrac{AB}{AC}$
    Mà AB < AC -> $\dfrac{AB}{AC}$ < 1
    => $\dfrac{MB}{MC}$ < 1 => MB < MC

    toan-lop-7-cho-abc-ab-ac-co-am-la-phan-giac-goc-a-m-thuoc-bc-tren-ac-lay-diem-d-sao-choad-ab-a-c

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222-9+11+12:2*14+14 = ? ( )