a) Ta có:
$\angle ABE = 90^\circ - \angle ABC = \angle ACF$
$\angle AEB = 90^\circ - \angle ACB = \angle AFC$
Vậy $\Delta ABE \sim \Delta ACF$ theo góc.

b) Ta có:
$\dfrac{HE}{HF} = \dfrac{BE}{CF}$
$\dfrac{HB}{HC} = \dfrac{BE}{CF}$
Nhân hai vế ta được: $HE \cdot HB = HF \cdot HC$

c) Ta có:
$\angle AEF = \angle AEB + \angle BEF = \angle ACF + \angle CBF = \angle ACB$
$\angle AFE = \angle AFC + \angle CBF = \angle ABC$
Vậy $\Delta AEF \sim \Delta ABC$ theo góc.