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 5 of 5: Conclude via the radical axis

Previous step Next step

AS\cdot AB=AT\cdot AC\implies A\in XY

Detailed analysis

The product $AS\cdot AB$ is the power of $A$ with respect to the circumcircle of $BQX$ (secant through $S,B$ ), and $AT\cdot AC$ is the power of $A$ with respect to the circumcircle of $CPX$ (secant through $T,C$ ). Their equality means $A$ lies on the radical axis of the two circles, which is exactly line $XY$ . Therefore $A$ , $X$ , $Y$ are collinear.