Lời giải:
a) Ta có:
$\triangle ABC$ cân tại $A\quad (gt)$
$OB = OC = R$
$\Rightarrow AO\perp BC$
Ta lại có: $AD$ là đường kính
$\Rightarrow A,O,D$ thẳng hàng
$\Rightarrow AD\perp BD$
$\Rightarrow \mathop{AB}\limits^{\displaystyle\frown}= \mathop{BD}\limits^{\displaystyle\frown}= \mathop{DC}\limits^{\displaystyle\frown}= \mathop{CA}\limits^{\displaystyle\frown}$
Do đó:
$\mathop{BE}\limits^{\displaystyle\frown}= \mathop{AB}\limits^{\displaystyle\frown}+ \mathop{AE}\limits^{\displaystyle\frown}= \mathop{BD}\limits^{\displaystyle\frown}+ \mathop{AE}\limits^{\displaystyle\frown}$
$\Rightarrow sđ\mathop{BE}\limits^{\displaystyle\frown}=sđ\mathop{BD}\limits^{\displaystyle\frown}+ sđ\mathop{AE}\limits^{\displaystyle\frown}$
$\Rightarrow \dfrac12sđ\mathop{BE}\limits^{\displaystyle\frown}=\dfrac12\left(sđ\mathop{BD}\limits^{\displaystyle\frown}+ sđ\mathop{AE}\limits^{\displaystyle\frown}\right)$
$\Rightarrow \widehat{ECB}=\widehat{BHD}$
$\Rightarrow HOCE$ là tứ giác nội tiếp
b) Ta có: $\widehat{BEC}= 90^\circ$ (góc nội tiếp chắn nửa đường tròn)
$\Rightarrow \widehat{BEC}=\widehat{BOH}= 90^\circ$
Xét $\triangle BHO$ và $\triangle BCE$ có:
$\begin{cases}\widehat{B}:\ \text{góc chung}\\\widehat{BEC}=\widehat{BOH}\quad (cmt)\end{cases}$
Do đó: $\triangle BHO\backsim \triangle BCE\ (g.g)$
$\Rightarrow\dfrac{BH}{BC}=\dfrac{BO}{BE}$
$\Rightarrow BH.BE = BO.BC = R.2R = 2R^2$
