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 1 of 5: Setup: two similar triangles at the base

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\beta := \angle QAC = \angle CBA, \quad \gamma := \angle PAB = \angle ACB \implies \triangle ABP \sim \triangle CAQ

Detailed analysis

Write $\beta=\angle QAC=\angle CBA$ and $\gamma=\angle PAB=\angle ACB$ as given. In triangle $ABP$ , the angle at $A$ is $\gamma$ and the angle at $B$ is $\angle ABC=\beta$ ; in triangle $CAQ$ , the angle at $C$ is $\gamma$ and the angle at $A$ is $\beta$ . Matching these equal angle pairs gives $\triangle ABP\sim\triangle CAQ$ (with $A\leftrightarrow C$ , $B\leftrightarrow A$ ), hence $\dfrac{BP}{PA}=\dfrac{AQ}{QC}$ .