1. In triangle ABC, the altitude to side BC, the perpendicular bisector of the arc $\mathrm{ABC}$, and the perpendicular bisector of side $\mathrm{AB}$ intersect at one point. Determine the measure of angle $\mathrm{ABC}$.
Solution: Let S be the intersection point of the two perpendicular bisectors and the altitude in $\triangle \mathrm{ABC}=\beta$. Since $\mathrm{S}$ lies on the perpendicular bisector of side $A B$, it follows that $\overline{\mathrm{SA}}=\overline{\mathrm{SB}}$, so $\triangle \mathrm{ABS}$ is isosceles, and $\triangle S A B=\triangle S B A=\frac{\beta}{2}$. From $\triangle \triangle B A_{1}: \frac{\beta}{2}+\beta=90^{\circ}, \beta=60^{\circ}$.

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