The circle $C_1$ of center $O$ and the circle $C_2$ intersect at points $A$ and $B$, so that point $O$ lies on circle $C_2$. The lines $d$ and $e$ are tangent at point $A$ to the circles $C_1$ and $C_2$ respectively. If the line $e$ intersects the circle $C_1$ at point $D$, prove that the lines $BD$ and $d$ are parallel.
Let the center of $C_2$ be $P$. Then let $\angle OAD=\angle ODA=\alpha$. Therefore, $\angle OAP=\angle OBP=90-\alpha$ and $\angle AOD=189-2\alpha$. Now extend $d$ until it intersects $C_2$ at $E$. Notice that $\angle DAE=90+\alpha$ so to prove $BD$ and $d$ are parallel, it suffices to prove that $\angle ADB=90-\alpha$. We will now let $\angle AEB=\beta$ so $\angle APB=2\beta$ and $\angle AOB=180-\beta$. Now look at quadrilateral $AOBP$. Since the angles in this add to $360$, $180+\beta=180+2\alpha$ so $\beta=2\alpha$. Now look at the angles around $O$. Since these must add to $360$, $\angle DOB=4\alpha$ so $\angle ODB=90-2\alpha$. Thus, $\angle ADB=90-\alpha$ as desired.