Problem 4

Let $ABCDE$ be a convex pentagon such that $BC=DE$. Assume that there is a point $T$ inside $ABCDE$ with $TB=TD,TC=TE$ and $\angle ABT = \angle TEA$. Let line $AB$ intersect lines $CD$ and $CT$ at points $P$ and $Q$, respectively. Assume that the points $P,B,A,Q$ occur on their line in that order. Let line $AE$ intersect $CD$ and $DT$ at points $R$ and $S$, respectively. Assume that the points $R,E,A,S$ occur on their line in that order. Prove that the points $P,S,Q,R$ lie on a circle.

Solution

First, we note the SSS congruence $\triangle TBC\cong\triangle TDE$. The key claim is the following:

Claim: $C,D,Q,S$ are concyclic.

Proof: Notice that $\triangle TBQ\sim\triangle TES$, so
$$\frac{TQ}{TS} = \frac{TB}{TE} = \frac{TD}{TC},$$implying the result. $\blacksquare$

Now the problem follows from
\begin{align*} \measuredangle PQS &= \measuredangle BQS \\ &= \measuredangle BQT + \measuredangle CQS \\ &= \measuredangle TSE + \measuredangle CDS \\ &= \measuredangle (CD, SE)\\ &= \measuredangle CRS. \end{align*}