If the altitude from one vertex of a triangle bisects the opposite side. Prove that the triangle is isosceles.

Circles $C_1$ and $C_2,$ of radii $r$ and $R$ respectively, touch each other as shown in the figure. The line $l$ which is parallel to the line joining the centres of $C_1$ and $C_2,$ is tangent to $C_1$ at $P$ and intersects $C_2$ at $A,B$. If $R^2=2r^2,$ then $\angle AOB$ equals-

In the given figure, side $BC$ of $∆ABC$ is bisected at $D$ and $O$ is any point on $AD. BO$ and $\mathrm{CO}$ produced meet $A\mathrm{C}$ and $AB$ at $E$ and $F$ respectively, and $AD$ is produced to $X$ so that $D$ is the midpoint of $OX.$ Prove that $AO:AX=AF:AB$ and show that $EF\parallel BC.$