Question on two intersecting circles at two points

Question

Two intersecting circles $S$ and $T$ in the plane intersect at two distinct points $A$ and $B$ and the center of $T$ lies on $S$. Points $C$ and $D$ are on $S$ and $T$ respectively with $C,B,D$ are collinear. Let point $E$ on $T$ such that $DE$ is parallel to $AC$. How do I show that $AE=AB$?

Answer 1

As mentioned in question, center of $T$ lies on $S$. Name it $O$. So we will have a cyclic quadrilateral $AOBC$. Then $$\overline{AO}=\overline{BO}=R_{\bigcirc T} \Longrightarrow \angle OAB = \angle OBA \tag{1}$$ By properties of cyclic quadrilaterals $$\angle OCB = \angle OAB \tag{2}$$ $$\angle OCA = \angle OBA \tag{3}$$ From $(1)$,$(2)$ and $(3)$ $$\angle OAB = \frac{\angle C}{2} \tag{♥}$$ There is another cyclic quadrilateral,$AEDB$.By properties of cyclic quadrilaterals $$\angle BAE+\angle D=180^\circ \tag{4}$$ Quadrilateral $ACDE$ is a trapezoid and $AC$ and $DE$ are it bases. So $$\angle C + \angle D = 180^\circ \tag{5}$$ From $(4)$ and $(5)$ $$\angle BAE = \angle C \tag{♦}$$Finally,From $(♥)$ and $(♦)$ $$\angle OAB =\angle OAE $$ Now it is easy to show $\triangle AOB$ and $\triangle AOE$ are congruent.So $AB=AE$.