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 6 of 7: Two circles pin

M

down as the Miquel point of all four lines

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In plain words

Four lines in general position always have a common Miquel point on all four of their triangles' circumcircles; being on two of those circles is already enough to identify which point that is.

M\in(\triangle RAF)\cap(\triangle QDF)\ \Longrightarrow\ M\ \text{is the Miquel point of lines}\ DA,EF,BD,AC

Detailed analysis

The four lines $DA,EF,BD,AC$ determine four triangles — $DFQ$ , $AFR$ , $APD$ (Step 1), and $PQR$ — and a classical theorem says their four circumcircles always share one common point, the Miquel point of the complete quadrilateral. Circles $(RAF)$ and $(QDF)$ both pass through $F$ and, by Steps 4–5, both pass through $M$ ; since two distinct circles meet in at most two points and the Miquel point lies on both, $M$ must be that Miquel point (it is not $F$ , which is one of the defining vertices, not the Miquel point).