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 1 of 6: Setup: second intersection

M

and its mirror

D'

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M=\odot(ADZ)\cap AB\ (M\ne A),\quad D'=\text{reflection of } D \text{ over } M

Detailed analysis

Let $N$ be the midpoint of $AC$ , and let $M$ be the second intersection of the circumcircle of $\triangle ADZ$ with segment $AB$ (so $M\ne A$ ). Let $D'$ be the reflection of $D$ in $M$ . It suffices to show $M$ is the midpoint of $AB$ , which follows once we show $ADBD'$ is a parallelogram.