VII.5. In the square $ABCD$, a point $P$ is chosen on side $BC$ and a point $S$ on side $CD$, such that $\triangle APB = \triangle APS$. Determine the measure of angle $PAS$.

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Solution: Let T be the foot of the perpendicular from vertex $\mathrm{A}$ to the segment PS. $\triangle \mathrm{ABP} \cong \triangle$ ATP (common hypotenuse and $\triangle A P B=\triangle A P T)$. It follows: $\triangle \mathrm{BAP}=\triangle \mathrm{PAT}=\mathrm{x} ; \triangle \mathrm{ATS} \cong \triangle A D S$ (common hypotenuse and equal legs). It follows $\triangle T A S=\triangle S A D=y ; 2 x+2 y=90^{\circ}$, $\mathrm{x}+\mathrm{y}=450$, hence $\triangle \mathrm{PAS}=450$.

![](https://cdn.mathpix.com/cropped/2024_06_05_71b0537f301960b82a41g-2.jpg?height=531&width=435&top_left_y=1643&top_left_x=917)