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 1 of 6: Get congruent triangles at

T

from the given equal lengths

\triangle TBC\cong\triangle TDE\ \Rightarrow\ \angle BTC=\angle DTE

Detailed analysis

Since $TB=TD$ , $TC=TE$ , and $BC=DE$ , triangles $TBC$ and $TDE$ are congruent by SSS, so $\angle BTC=\angle DTE$ .