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 3 of 5: Equal angles at M and C

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\angle BMP = \angle NCQ

Detailed analysis

From the similarity $\triangle BPM\sim\triangle NQC$ established above, corresponding angles give $\angle BMP=\angle NCQ$ , i.e. $\angle BMP=\angle NCB$ .