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 6 of 6: Finish the angle chase to conclude

PQRS

is cyclic

Previous step Next step

\angle QCD-\angle PQC=\angle QSD-\angle EST=\angle QSR\ \Rightarrow\ PQRS\ \text{cyclic}

Detailed analysis

Substituting $\angle QCD=\angle QSD$ and $\angle PQC=\angle EST$ from the previous steps, $\angle QCD-\angle PQC=\angle QSD-\angle EST$ ; since $S,T,D$ are collinear and $S,E,A,R$ are collinear, $\angle EST$ measures the same ray configuration as $\angle DSE$ , so $\angle QSD-\angle EST=\angle QSR$ . Combining with the previous step, $\angle QPR=\angle QSR$ , so $P$ and $S$ see segment $QR$ at equal angles, which means $P,Q,R,S$ lie on a common circle.