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 2 of 6: Turn this into equal angles at

Q

and

S

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\angle BTQ=180^\circ-\angle BTC=180^\circ-\angle DTE=\angle STE

Detailed analysis

Since $Q$ lies on line $CT$ beyond $T$ from $C$ and $S$ lies on line $DT$ beyond $T$ from $D$ , $\angle BTQ=180^\circ-\angle BTC=180^\circ-\angle DTE=\angle STE$ .