Prove that AA" ,CC" is perpendicular to bisector of B

Question

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)

Answer 1

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.

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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.

Answer 2

\begin{align} \angle BA_2C_1+\angle C_1BA_2 &= \angle A_1C_2B+\angle C_2BA_1= \beta ,\\ \angle BA_2C_1&=\angle C_1BA_2 = \angle A_1C_2B=\angle C_2BA_1= \tfrac\beta2 . \end{align}

Hence, triangles $C_1BA_2$ and $A_1C_2B$ are isosceles,

\begin{align} |C_1A|&=|C_1B|=|C_1A_2| ,\\ |A_1B|&=|A_1C|=|A_1C_2| , \end{align}

$A_1$ and $C_1$ are the centers of the circles with diameters $|AB|$ and $|BC|$, hence $\angle BA_2A=\angle CC_2B=90^\circ$.