Problem 5

Line

\ell

intersects sides

BC

and

AD

of cyclic quadrilateral

ABCD

at its interior points

R

and

S

, respectively, and intersects ray

DC

beyond point

C

Q

, and ray

BA

beyond point

A

P

. The circumcircles of triangles

QCR

and

QDS

intersect at

N\neq Q

, while the circumcircles of triangles

PAS

and

PBR

intersect at

M\neq P

. Let lines

MP

and

NQ

meet at point

X

, lines

AB

and

CD

meet at point

K

, and lines

BC

and

AD

meet at point

L

. Prove that point

X

lies on line

KL

Step 4 of 6: MP meets KL on the same circle

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V=MP\cap KL\implies V\in(EMN)

Detailed analysis

Symmetrically, let $V=MP\cap KL$ . Using the circles $(LCD)$ (which contains $E$ and $N$ ), $(CQR)$ (which also contains $N$ ), and $(MNQP)$ , the analogous directed-angle computation shows $V$ lies on $\omega=(EMN)$ as well, or that $KL$ is tangent to $\omega$ at $E$ if $V=E$ .