Question –
The internal bisector of the $\angle B$ of $\Delta ABC$ meets the sides $B'C'$ and $B'A'$ of the medial triangle in the points $A''$ and $C''$ respectively. Prove that $AA''$ and $CC''$ are perpendicular to the bisector of $\angle B$ and that $B'A''=B'C''$. (Similar results holds for external bisector)
Figure – 
My attempt –
First I tried to make $\angle AA''B = \angle AA'B$ so that they will be in same segment $AB$ so they will be concyclic and angle in semi-circle is $90^o$ so I will get result.
But even after spending sufficient amount of time I am not able to show that those angles are equal. I have tried angle bisector theorem, similarity, etc but none of them worked out.
Any hint will be appreciated.
Thank you in advance.
Source: CTPCM (Olympiad book)
Best Answer
Draw X on BC such that A''X // AB. Note that the green triangle is congruent to the blue triangle (by AAS). In addition, each of these triangles are isoscelss because their base angles are equal.
Then, BB' = B'A". Since Ac' = C' B already because of the median trianlge. C' is then the center of of the circle with radius C'A = C'B = C'A". that circle will pass through A, B, A''. This means $\angle AA"B = 90^0$. This further means AA" is perpendicular to the angle bisector of ∠ABC.
In our usual naming convention, A' is a point on the side opposing A. That is, A' is on BC. Similarly B' is on AC; and C' is on AB.