9.3. Inside an isosceles triangle $\mathrm{ABC}$ with equal sides $\mathrm{AB}=\mathrm{BC}$ and an angle of 80 degrees at vertex $\mathrm{B}$, a point $\mathrm{M}$ is taken such that the angle

$\mathrm{MAC}$ is 10 degrees, and the angle $\mathrm{MCA}$ is 30 degrees. Find the measure of angle $\mathrm{AMB}$.
Answer: 70 degrees.

Solution. Draw a perpendicular from vertex B to side AC, and denote the points of its intersection with lines AC and CM as P and T, respectively. Since angle MAC is less than angle MCA, side CM of triangle MAC is shorter than side AM, so point M is closer to C than to A, and therefore T lies on the extension of segment CM. Point T lies on the perpendicular bisector AP of segment AC, so triangle ATC is isosceles, hence angle TAC is 30 degrees, and therefore angles BAT and MAT are 20 degrees each. The measures of angles $\mathrm{ABP}=\mathrm{ABC} / 2=80 / 2=40$ and $\mathrm{AMT}=\mathrm{MAC}+\mathrm{MCA}=10+30=40$ are equal, so triangles ABT and AMT are congruent by the common side AM and the adjacent angles. Therefore, the corresponding sides $\mathrm{AB}$ and $\mathrm{AM}$ are equal, and triangle $\mathrm{AMB}$ is isosceles. Consequently, its angle AMB at the base MB is $(180- BAM) / 2=(180-(50-10)) / 2=(180-40) / 2=70$ degrees.

Grading criteria. Proof that T lies on the extension of segment CM: 1 point. Proof of the isosceles nature of triangle ATC: 1 point. Proof of the isosceles nature of triangle AMB: 4 points. Finding the angle AMB: 1 point.