Lời giải:
Xét $\triangle AHE$ và $\triangle ACD$ có:
$\begin{cases}\widehat{A}:\ \text{góc chung}\\\widehat{E} = \widehat{D} = 90^\circ\end{cases}$
Do đó: $\triangle AHE\backsim \triangle ACD\ (g.g)$
$\Rightarrow \dfrac{AH}{AC} = \dfrac{AE}{AD}$
$\Rightarrow AH.AD = AE.AC\qquad (1)$
Ta có:
$\widehat{BEC} = \widehat{BFC} = 90^\circ$
$\Rightarrow BCEF$ là tứ giác nội tiếp
$\Rightarrow \widehat{AEF} = \widehat{ABC}$
$\Rightarrow \widehat{AEM} = 180^\circ - \widehat{AEF} = 180^\circ - \widehat{ABC}$
$\Rightarrow \widehat{AEM} = 180^circ - \dfrac12sđ\mathop{AC}\limits^{\displaystyle\frown}$
$\Rightarrow \widehat{AEM} = \dfrac12\left(360^circ - sđ\mathop{AC}\limits^{\displaystyle\frown}\right)$
$\Rightarrow \widehat{AEM} = \dfrac12sđ\mathop{ABC}\limits^{\displaystyle\frown}$
$\Rightarrow \widehat{AEM} = \widehat{AMC}$
Xét $\triangle AEM$ và $\triangle AMC$ có:
$\begin{cases}\widehat{A}:\ \text{góc chung}\\\widehat{AEM} = \widehat{AMC}\quad (cmt)\end{cases}$
Do đó: $\triangle AEM\backsim\triangle AMC\ (g.g)$
$\Rightarrow \dfrac{AE}{AM} = \dfrac{AM}{AC}$
$\Rightarrow AM^2 = AE.AC\qquad (2)$
Từ $(1)(2)\Rightarrow AM^2 = AH.AD$
