Ta có:
$\begin{cases}\widehat{HAC}+\widehat{A_3}=90^o\\\widehat{B}+\widehat{A_3}=90^o\end{cases}$
$⇒ \widehat{B}=\widehat{HAC}$
$⇒ \dfrac{1}{2}\widehat{B}=\dfrac{1}{2}\widehat{HAC}$
$⇒ \dfrac{1}{2} [180^o-\widehat{2H_1}]=\widehat{A_1}$
$⇒ 90^o-\widehat{H_1}=\widehat{A_1}$
$⇒ \widehat{H_2}=\widehat{A_1}$
Mà 2 góc ở vị trí so le trong $⇒ EH//AD$