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 4 of 6: Congruent triangles give

D'A=DB

Previous step Next step

\triangle D'ZA\cong\triangle DZC\ (SAS) \implies D'A=DC=DB

Detailed analysis

Since $\angle D'ZA=\angle DZC$ , $ZD'=ZD$ , and $ZA=ZC$ , triangles $D'ZA$ and $DZC$ are congruent by $SAS$ . Hence $D'A=DC$ . Since $D$ is the midpoint of arc $BC$ not containing $A$ (as $AD$ bisects $\angle A$ ), $DB=DC$ ; therefore $D'A=DB$ .