Problem 2

Let

ABC

be a triangle with

AB<AC

. Let

D

be the intersection point of the internal bisector of

\angle BAC

and the circumcircle of

ABC

. Let

Z

be the intersection point of the perpendicular bisector of

AC

with the external bisector of

\angle BAC

. Prove that the midpoint of segment

AB

lies on the circumcircle of triangle

ADZ

Step 3 of 6: Angle computation via the cyclic quadrilateral

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\alpha:=\angle BAD=\angle DAC,\quad \angle MZD=\angle MAD=\alpha,\ \angle AZC=2\alpha \implies \angle D'ZA=\angle DZC

Detailed analysis

Let $\alpha=\angle BAD=\angle DAC$ . Since $AZDM$ is cyclic, $\angle MZD=\angle MAD=\alpha$ , so $\angle D'ZD=2\alpha$ (as $M$ is the midpoint of $DD'$ , giving $\angle D'ZM=\angle MZD$ ). Also, using that $Z$ lies on the perpendicular bisector of $AC$ together with $\angle DAC=\alpha$ , one finds $\angle AZC=2\alpha$ . Subtracting $\angle AZD$ from both $\angle D'ZD$ and $\angle AZC$ gives $\angle D'ZA=\angle D'ZD-\angle AZD=\angle AZC-\angle AZD=\angle DZC$ .