Giải thích các bước giải:

1.Xét $\Delta ADH,\Delta AHB$ có:

Chung $\hat A$

$\widehat{ADH}=\widehat{AHB}(=90^o)$

$\to \Delta ADH\sim\Delta AHB(g.g)$

$\to \dfrac{AD}{AH}=\dfrac{AH}{AB}$

$\to AH^2=AD.AB$

Tương tự: $AE.AC=AH^2$

2.Xét $\Delta ADE,\Delta ABC$ có:

Chung $\hat A$

$\dfrac{AD}{AC}=\dfrac{AE}{AB}$ vì $AD.AB=AE.AC(=AH^2)$

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

$\to \widehat{ADE}=\hat C=90^o-\widehat{EHC}=\widehat{AHE}$
3.Đề sai