Problem 1

Let

ABC

be an acute triangle. Let

D

be a point on side

AB

and

E

be a point on side

AC

such that lines

BC

and

DE

are parallel. Let

X

be an interior point of quadrilateral

BCED

. Suppose rays

DX

and

EX

meet side

BC

at points

P

and

Q

, respectively, such that both

P

and

Q

lie between

B

and

C

. Suppose that the circumcircles of triangles

BQX

and

CPX

intersect at a point

Y\neq X

. Prove that points

A

X

, and

Y

are collinear.

Step 2 of 5: Put T on the same circle

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T=(CPX)\cap AC,\ T\neq C

Detailed analysis

Symmetrically, let the circumcircle of $CPX$ meet line $AC$ again at $T$ . Since $DE\parallel BC$ , the transversal line $DXP$ gives $\angle EDX=\angle XPB$ , and since $C,P,X,T$ are concyclic with $B$ on ray $CP$ beyond $P$ , the exterior angle $\angle XPB$ equals the opposite interior angle $\angle XTE$ . Hence $\angle EDX=\angle XTE$ , so $T$ also lies on the circle through $D,E,X$ .