Lời giải:
a) Xét $\triangle AHB$ và $\triangle CAB$ có:
$\begin{cases}\widehat{H} = \widehat{A} = 90^\circ\\\widehat{B}:\ \text{góc chung}\end{cases}$
Do đó $\triangle AHB\backsim \triangle CAB\ (g.g)$
$\Rightarrow \dfrac{AB}{BC} = \dfrac{BH}{AB}$
$\Rightarrow AB^2 = BH.BC$
b) Ta có: $\triangle AHB\backsim \triangle CAB$ (câu a)
$\Rightarrow \dfrac{AH}{AC} = \dfrac{BH}{AB}$
$\Rightarrow \dfrac{2AQ}{AC} = \dfrac{2BP}{AB}$
$\Rightarrow \dfrac{AQ}{BP} = \dfrac{AC}{AB}$
Xét $\triangle AQC$ và $\triangle BPA$ có:
$\begin{cases}\dfrac{AQ}{BP} = \dfrac{AC}{AB}\quad (cmt)\\\widehat{QAC} = \widehat{PBA}\quad \text{(cùng phụ $\widehat{HAB}$)}\end{cases}$
Do đó $\triangle AQC\backsim \triangle BPA\ (c.g.c)$
c) Ta có: $\triangle AQC\backsim \triangle BPA$ (câu b)
$\Rightarrow \widehat{ACQ} = \widehat{BAP}$
Khi đó:
$\quad \widehat{APC} + \widehat{PCQ}$
$= \widehat{PBA} + \widehat{BAP} + \widehat{HCQ}$
$= \widehat{QAC} + \widehat{ACQ} + \widehat{HCQ}$
$= \widehat{HAC} + \widehat{HCA}$
$= 90^\circ$
Do đó $AP\perp CQ$
