Problem 1

Let

\Gamma

be the circumcircle of triangle

ABC

. Let

D

be a point on side

BC

. The tangent to

\Gamma

A

intersects the line through

D

parallel to

BA

at point

E

. The segment

CE

intersects

\Gamma

again at

F

. Suppose

B

D

F

E

are concyclic. Prove that

AC

BF

DE

are concurrent.

Step 2 of 4: Parallel line gives the same corresponding angle

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\angle EDC=\angle ABC \implies \angle EAC=\angle EDC

Detailed analysis

Line $BC$ is a transversal cutting the parallel lines $AB$ and $DE$ at $B$ and $D$ , so the corresponding angles are equal: $\angle EDC=\angle ABC$ . Combined with the previous step, $\angle EAC=\angle EDC$ , so $A$ and $D$ subtend equal angles on segment $EC$ and lie on the same side of it; hence $A$ , $D$ , $C$ , $E$ are concyclic.