Problem 5

Let

ABCD

be a fixed convex quadrilateral with

BC = DA

and

BC \nparallel DA

. Let two variable points

E

and

F

lie on the sides

BC

and

DA

, respectively, and satisfy

BE = DF

. The lines

AC

and

BD

meet at

P

, the lines

BD

and

EF

meet at

Q

, the lines

EF

and

AC

meet at

R

. Prove that the circumcircles of the triangles

PQR

, as

E

and

F

vary, have a common point other than

P

Step 4 of 7:

M

lies on the circle through

R,A,F

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In plain words

Applying the spiral-similarity lemma again, now to the pair $C\mapsto A$ , $E\mapsto F$ under the very same rotation, plants $M$ on a new circle through the point where lines $CA$ and $EF$ cross.

R=EF\cap AC\ \Longrightarrow\ M\in(\triangle RAF)

Detailed analysis

Since the fixed rotation about $M$ sends $C\mapsto A$ and $E\mapsto F$ , the spiral-similarity lemma (applied to this pair) places $M$ on the circumcircle of the triangle formed by $R=CA\cap EF$ together with $A$ and $F$ . So $M\in(\triangle RAF)$ .