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 3 of 6: NQ meets KL on the circle through E, M, N

Previous step Next step

T=NQ\cap KL\implies T\in(EMN)

Detailed analysis

Let $T=NQ\cap KL$ . Using the circles $(LAB)$ (which contains $E$ and $M$ by definition of the Miquel point), $(APS)$ (which also contains $M$ ), and $(MNQP)$ from Step 1, directed angles give $\angle TEM=\angle LEM=\angle LAM=\angle SAM=\angle SPM=\angle QPM=\angle QNM=\angle TNM$ . Hence $T$ lies on the circle $\omega=(EMN)$ , or $KL$ is tangent to $\omega$ at $E$ if $T=E$ .