Problem 9.3. Point $I$ is the intersection of the angle bisectors of triangle $ABC$. On side $BC$, point $X$ is marked. It turns out that $AI = BX, AC = CX, \angle ABC = 42^\circ$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f585fb32d0f8951822d4g-3.jpg?height=293&width=450&top_left_y=580&top_left_x=502)

(a) (2 points) How many degrees is angle $BIX$?

(b) (2 points) How many degrees is angle $BCA$?
# Answer:

(a) (2 points) $21^{\circ}$.

(b) (2 points) $54^{\circ}$.

Solution. (a) Note that triangles XIC and AIC are congruent (IC is a common side, $X C = C A, \angle X C I = \angle A C I$ by the problem statement; see Fig. 3). Therefore, $X I = A I = B X$, and triangle $B X I$ is isosceles. Hence,

$$
\angle B I X = \angle I B X = \frac{\angle A B C}{2} = 21^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f585fb32d0f8951822d4g-4.jpg?height=331&width=515&top_left_y=412&top_left_x=469)

Fig. 3: to the solution of problem 9.3

(b) It follows that $\angle A B I = \angle B I X = 21^{\circ}$, i.e., $A B \| X I$. It remains to find the angles $\angle B A C$ and $\angle B C A$.

$$
\begin{gathered}
\angle B A C = 2 \angle I A C = 2 \angle I X C = 2 \angle A B C = 84^{\circ} \\
\angle A C B = 180^{\circ} - \angle B A C - \angle A B C = 54^{\circ} .
\end{gathered}
$$