Problem 5

An arbitrary point

M

is selected in the interior of segment

AB

. Squares

AMCD

and

MBEF

are constructed on the same side of

AB

, with circumcenters

P

and

Q

. Their circumcircles meet again at

N

. Let

N'

be the intersection of

AF

and

BC

. (a) Prove

N=N'

. (b) Prove that

MN

passes through a fixed point independent of

M

. (c) Find the locus of the midpoint of

PQ

M

varies.

Step 1 of 5: Use the congruent right triangles

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In plain words

The two squares turn the pieces around $M$ into congruent right triangles.

\triangle AFM\cong\triangle CBM\quad\Longrightarrow\quad\angle AN'B=90^\circ

Detailed analysis

Because $AM=CM$ and $MB=MF$ , with the right angles supplied by the squares, triangles $AFM$ and $CBM$ are congruent. Thus the angles they make with $AB$ are complementary, so at $N'=AF\cap BC$ we have $\angle AN'B=90^\circ$ .