Lời giải:
a) Xét $\triangle ABD$ và $\triangle ACE$ có:
$\begin{cases}\widehat{A}:\ \text{góc chung}\\\widehat{D} = \widehat{E} = 90^\circ\end{cases}$
Do đó $\triangle ABD\backsim \triangle ACE\ (g.g)$
$\Rightarrow \dfrac{AD}{AE} = \dfrac{AB}{AC}$
$\Rightarrow \dfrac{AD}{AB} = \dfrac{AE}{AC}$
Xét $\triangle ADE$ và $\triangle ABC$ có:
$\begin{cases}\dfrac{AD}{AB} = \dfrac{AE}{AC}\quad (cmt)\\\widehat{A}:\ \text{góc chung}\end{cases}$
Do đó $\triangle ADE\backsim \triangle ABC\ (c.g.c)$
b) Ta có: $\triangle ADE\backsim \triangle ABC$ (câu a)
$\Rightarrow \dfrac{ED}{BC} = \dfrac{AD}{AB}$
$\Rightarrow \dfrac{ED}{BC} = \cos\widehat{DAB}$
$\Rightarrow ED = BC.\cos\widehat{DAB} = 4.\cos60^\circ$
$\Rightarrow ED = 2\ cm$
c) Xét $\triangle ADH$ và $\triangle BDC$ có:
$\begin{cases}\widehat{DAH} = \widehat{HBC}\quad \text{(cùng phụ $\widehat{C}$)}\\\widehat{ADH} = \widehat{BDC} = 90^\circ\end{cases}$
Do đó $\triangle ADH\backsim \triangle BDC\ (g.g)$
$\Rightarrow \dfrac{AH}{BC} = \dfrac{AD}{BD}$
$\Rightarrow \dfrac{AH}{BC} =\cot\widehat{DAB}$
$\Rightarrow AH = BC.\cot\widehat{DAB} = 4.\cot60^\circ$
$\Rightarrow AH = \dfrac{4\sqrt3}{3}\ cm$
