How to find the angle by a triangle and a median and bisector?

Question

The problem is as follows:

In a triangle $\triangle{ABC}$, the measure of angle $\angle ABC = \angle 105^{\circ}$. On $AC$ it is located a point $M$, such that $AB = MC$. The mediatrices of $BC$ and $AM$ intersect at $Q$. Find the measure of $\angle BAC$, if the measure of $\angle BCA = \angle ACQ$.

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&30^{\circ}\\ 2.&40^{\circ}\\ 3.&45^{\circ}\\ 4.&35^{\circ}\\ 4.&50^{\circ}\\ \end{array}$

Can someone help me with the right interpretation of this problem?. I'm having difficulties with the right drawing.

I found this problem in my book in the chapter of triangle congruency thus I believe the approach for this problem is intended to follow such route. Can someone help me with that?. Please include a drawing in the answer so I can understand. I'm lost if this requires some sort of construction. Please try to include a solution explained step-by-step.

Answer 1

Before we dig into this question, note that I don't draw the mediatrices to avoid making the picture too complicated.

Connect $\overline{BQ}$. Let $\angle BCA=\angle ACQ=\theta$.

Since $Q$ is on the perpendicular bisector of $\overline{BC}$, we have $\color{blue}{\overline{BQ}=\overline{CQ}}$.

Since $Q$ is on the perpendicular bisector of $\overline{AM}$, we have $\color{red}{\overline{AQ}=\overline{MQ}}$.

Also from the question we already know $\color{purple}{\overline{AB}=\overline{MC}}$.

Hence $$\triangle BAQ\cong\triangle CMQ\quad(S.S.S.)$$ so $$\angle ABQ=\angle MCQ=\theta$$

We also have $$\angle QBC=\angle QCB=2\theta$$ since $\triangle BCQ$ is an isosceles. Therefore we can say $$\theta+2\theta=105^{\circ}\implies \theta =35^{\circ}$$

Finally $$\angle BAC=180^{\circ}-105^{\circ}-\theta=\color{red}{40^{\circ}}$$