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 6 of 6: Conclude:

ADBD'

is a parallelogram

Previous step Next step

AD'\parallel BD,\ AD'=BD \implies ADBD' \text{ parallelogram} \implies M=\text{midpoint of } AB

Detailed analysis

Since $AD'\parallel BD$ and $AD'=DB$ (shown above), quadrilateral $ADBD'$ is a parallelogram. Its diagonals $AB$ and $DD'$ bisect each other, and $M$ is by construction the midpoint of $DD'$ ; hence $M$ is also the midpoint of $AB$ , which is exactly the claim.