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 2 of 7:

M

is a rotation center for

B\mapsto D

C\mapsto A

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In plain words

The classical spiral-similarity lemma converts the two circles through $M$ into a single spiral similarity centered at $M$ ; equal lengths $BC=DA$ upgrade that spiral similarity to a pure rotation.

M\ \text{is the center of the rotation sending}\ B\mapsto D,\ C\mapsto A

Detailed analysis

By the spiral-similarity lemma, since circles $(APD)$ and $(BPC)$ meet at $P$ and $M$ , the point $M$ is the center of the spiral similarity sending $B\mapsto D$ and $C\mapsto A$ (equivalently, sending segment $BC$ to segment $DA$ ). Because $BC=DA$ , the ratio of this spiral similarity is $1$ , so it is in fact a pure rotation about $M$ .