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 5 of 6: A line meets a circle in at most two points

Previous step Next step

\omega\cap KL=\{T,V\}\text{ or tangent at }E\implies T=V

Detailed analysis

The line $KL$ meets the circle $\omega=(EMN)$ in at most two points. By Steps 3 and 4, both $T$ and $V$ are among these intersection points (with $E$ itself being one of them, possibly as a point of tangency). If $T\neq E$ and $V\neq E$ , they must be the same second intersection point, so $T=V$ ; if either equals $E$ , the tangency forces the other to equal $E$ too. In every case $T=V$ .