Ai làm được không nhỉ

$\textit{Answer:}$

$\text{a,}$ Xét $\triangle$$\text{ABD}$ và $\triangle$$\text{EBD}$ có:

$\text{AB}$ = $\text{BE}$ ( gt ).

$\widehat{ABD}$ = $\widehat{EBD}$ ( do $\text{BD}$ là tia phân giác $\widehat{ABD}$ ).

$\text{BD}$ chung.

$\Rightarrow$ $\triangle$$\text{ABD}$ = $\triangle$$\text{EBD}$ (c.g.c).

$\text{b,}$ Từ $\triangle$$\text{ABD}$ = $\triangle$$\text{EBD}$ (c/m a).

$\Rightarrow$ $\text{DE}$ = $\text{DA}$ ( cạnh t/ư ) và $\widehat{BED}$ = $\widehat{BAD}$ = $90^0$ ( góc t/ư ).

$\Rightarrow$ $\text{DE}$ $\bot$ $\text{BC}$.

$\text{c,}$ Từ $\text{DE}$ = $\text{DA}$ (c/m b).

Xét $\triangle$$\text{ADF}$ và $\triangle$$\text{EDC}$ có:

$\text{DE}$ = $\text{DA}$ (c/mt)

$\widehat{FAD}$ = $\widehat{DEC}$ = $90^0$

$\text{AF}$ = $\text{CE}$ (gt).

$\Rightarrow$  $\triangle$$\text{ADF}$ = $\triangle$$\text{EDC}$ (c.g.c)

$\Rightarrow$ $\widehat{ADF}$ = $\widehat{CDF}$

Mà $\text{A.D,C}$ thẳng hàng $\Rightarrow$ $\text{FDE}$ thẳng hàng.

$\textit{#minhthu5626}$

$\textit{#Math}$