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 2 of 5: Doubling to M, N preserves the similarity

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\frac{BP}{PM}=\frac{BP}{PA}=\frac{AQ}{QC}=\frac{NQ}{QC}, \quad \angle BPM = \beta+\gamma = \angle CQN

Detailed analysis

Since $P$ is the midpoint of $AM$ , $PM=PA$ , so the ratio from the previous step reads $\dfrac{BP}{PM}=\dfrac{AQ}{QC}=\dfrac{NQ}{QC}$ (using $NQ=QA$ ). Also $\angle BPM=180^\circ-\angle APB=\beta+\gamma$ (exterior angle of $\triangle ABP$ ), and likewise $\angle CQN=\beta+\gamma$ . Hence $\triangle BPM\sim\triangle NQC$ by SAS similarity.