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 4 of 5: Compare with the ratio from DE ∥ BC

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\frac{AD}{AB}=\frac{AE}{AC}\implies AS\cdot AB=AT\cdot AC

Detailed analysis

Because $DE\parallel BC$ , triangle $ADE$ is similar to triangle $ABC$ , so $AD/AB=AE/AC$ . Multiplying $AD\cdot AS=AE\cdot AT$ by the equal ratios $AB/AD=AC/AE$ gives $AS\cdot AB=AT\cdot AC$ .