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, with

P,B,A,Q

in that order on the line. Let line

AE

intersect lines

CD

and

DT

at points

R

and

S

, respectively, with

R,E,A,S

in that order on the line. Prove that the points

P

S

Q

R

lie on a circle.

Step 3 of 6: Combine with the angle hypothesis to get similar triangles

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\triangle TQB\sim\triangle TSE\ \Rightarrow\ \angle PQC=\angle EST,\quad \frac{QT}{ST}=\frac{TB}{TE}

Detailed analysis

Since $Q$ lies on ray $BA$ beyond $A$ and $S$ lies on ray $EA$ beyond $A$ , $\angle TBQ=\angle ABT$ and $\angle TES=\angle AET$ , so the hypothesis $\angle ABT=\angle AET$ gives $\angle TBQ=\angle TES$ . Combined with $\angle BTQ=\angle ETS$ from the previous step, $\triangle TQB\sim\triangle TSE$ by AA, which yields $\angle PQC=\angle EST$ and $QT/ST=TB/TE$ .