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 2 of 6:

ZD

is a diameter, giving

ZD'=ZD

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\angle ZAD=90^\circ \implies ZD \text{ diameter of } \odot(AZD) \implies \angle ZMD=90^\circ \implies ZD'=ZD,\quad ZA=ZC

Detailed analysis

Since the internal and external bisectors at $A$ are perpendicular, $\angle ZAD=90^\circ$ , so $ZD$ is a diameter of the circumcircle of $\triangle AZD$ , giving $\angle ZMD=90^\circ$ . As $M$ is the midpoint of $DD'$ by construction, $ZM\perp DD'$ makes $ZM$ the perpendicular bisector of $DD'$ , so $ZD'=ZD$ . Also $ZA=ZC$ since $Z$ lies on the perpendicular bisector of $AC$ .