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 1 of 5: Put S on the circle through D, E, X

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S=(BQX)\cap AB,\ S\neq B

Detailed analysis

Let the circumcircle of $BQX$ meet line $AB$ again at $S$ . Since $DE\parallel BC$ , the transversal line $EXQ$ gives $\angle DEX=\angle XQC$ (alternate angles). Because $B,Q,X,S$ are concyclic and $C$ lies on ray $BQ$ beyond $Q$ , the exterior angle $\angle XQC$ equals the opposite interior angle $\angle XSD$ . Hence $\angle DEX=\angle XSD$ , so $D,E,X,S$ are concyclic.