Cho Hình 14 Chứng minh tam giác ABD đồng dạng với tam giác CEB

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Cho Hình 14
Chứng minh tam giác ABD đồng dạng với tam giác CEB

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    Ta có: $\widehat{DBA}+\widehat{DBE}+\widehat{EBC}=180^o$

    Mà: $\widehat{DBE}=90^o$

    `⇒` $\widehat{DBA}+\widehat{EBC}=90^o$

    `⇒` $90^o-\widehat{ADB}+\widehat{EBC}=90^o$

    `⇒` $\widehat{EBC}-\widehat{ABD}=90^o-90^o=0$

    `⇒` $\widehat{EBC}=\widehat{ABD}$ 

    Xét `ΔABD` và `ΔCBE` có: 

    $\widehat{EBC}=\widehat{ABD}$ `(cmt)`

    $\widehat{DAB}=\widehat{ECB}(=90^o)$

    `⇒ ΔABD` ∼ `ΔCBE (g.g)`

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