'Carpenter's square' seems an odd name for something L shaped but the key to its name, as to this tough nut, is in right angles. To prove that $\angle \mathrm{PBQ}=\angle \mathrm{QBR}=\angle \mathrm{RBC}$ first draw the diagram with the carpenter's square placed so that R is on the line $\mathrm{DE}$, $P$ is on the line $\mathrm{AB}$ and the edge $\mathrm{QT}$ contains the point $B$. Draw the lines $\mathrm{BQ}$ and $\mathrm{BR}$. Then draw the line $\mathrm{RX}$ perpendicular to $\mathrm{BC}$ with $X$ on $\mathrm{BC}$ and look for three congruent triangles.