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 4 of 6: Use

TB=TD

TC=TE

to show

C,D,Q,S

concyclic

Previous step Next step

QT\cdot TE=QT\cdot TC=ST\cdot TB=ST\cdot TD\ \Rightarrow\ CDQS\ \text{cyclic}

Detailed analysis

Cross-multiplying $QT/ST=TB/TE$ gives $QT\cdot TE=ST\cdot TB$ ; substituting $TE=TC$ and $TB=TD$ gives the chain $QT\cdot TE=QT\cdot TC=ST\cdot TB=ST\cdot TD$ , so in particular $QT\cdot TC=ST\cdot TD$ . Since $T=CQ\cap DS$ , this is exactly the power-of-a-point criterion for $C,D,Q,S$ to be concyclic.