Problem 4

Points

P

and

Q

lie on side

BC

of an acute-angled triangle

ABC

so that

\angle PAB=\angle BCA

and

\angle CAQ=\angle ABC

. Points

M

and

N

lie on lines

AP

and

AQ

, respectively, such that

P

is the midpoint of

AM

, and

Q

is the midpoint of

AN

. Prove that the intersection of lines

BM

and

CN

lies on the circumcircle of triangle

ABC

Step 4 of 5: A second similarity centred at the intersection S

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S := BM \cap CN \implies \triangle BPM \sim \triangle BSC

Detailed analysis

Let $S$ be the intersection of lines $BM$ and $CN$ . Since $\angle BMP=\angle BMS$ equals $\angle SCB$ (rewriting $\angle NCB$ using that $S$ lies on line $CN$ ), and $\angle MBP=\angle SBC$ trivially (same line $BM$ ), triangles $BPM$ and $BCS$ share two equal angles, so $\triangle BPM\sim\triangle BSC$ .