sosssss cứu với ạ

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sosssss cứu với ạ

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    Giải thích các bước giải:

    a.Xét $\Delta DBH,\Delta DAC$ có:

    $\widehat{BDH}=\widehat{ADC}(=90^o)$

    $\widehat{BHD}=90^o-\widehat{HBD}=90^o-\widehat{EBC}=\widehat{ECB}=\widehat{ACD}$

    $\to \Delta DBH\sim\Delta DAC(g.g)$

    $\to \dfrac{DB}{DA}=\dfrac{HD}{DC}$

    $\to DB\cdot DC=DH\cdot DA$

    b.Vì $BC$ là đường kính của $(O)\to \widehat{BIC}=90^o\to \Delta IBC$ vuông tại $I$

    Mà $ID\perp BC$

    $\to DI^2=DB\cdot DC=DA\cdot DH=36$

    $\to DI=6$

    $\to AI=AD-DI=3$

    c.Xét $\Delta AEB,\Delta AFC$ có:

    Chung $\hat A$

    $\hat E=\hat F(=90^o)$

    $\to \Delta AEB\sim\Delta AFC(g.g)$

    $\to \dfrac{AE}{AF}=\dfrac{BA}{AC}$

    Mà $\widehat{FAE}=\widehat{BAC}$

    $\to \Delta AEF\sim\Delta ABC(c.g.c)$

    $\to \dfrac{EF}{BC}=\dfrac{AE}{AB}=\cos\hat A=\dfrac12$

    $\to EF=\dfrac12BC=\dfrac12a$

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