Let the excircle of triangle APQ opposite to the vertex A have center C' and touch the sideline AP of the triangle at the point B'. Then, $C^{\prime}B^{\prime}\perp AP$, and the distance AB' equals the semiperimeter of triangle APQ. Since the perimeter of triangle APQ equals 2, its semiperimeter must be 1; thus, we have AB' = 1. On the other hand, we know that AB = 1 (in fact, the square ABCD has sidelength 1). Since the points B and B' both lie on the ray AP, we thus can conclude that the points B and B' coincide. Hence, $C^{\prime}B^{\prime}\perp AP$ becomes $C^{\prime}B\perp AP$; in other words, $C^{\prime}B\perp AB$. Hence, the point C' lies on the perpendicular to the line AB at the point B. But since ABCD is a square, the perpendicular to the line AB at the point B is the line BC. Hence, the point C' lies on the line BC. Similarly, the point C' lies on the line DC. As the lines BC and DC have only one common point, namely the point C, it follows that the point C' coincides with the point C. Since we have defined the point C' as the center of the excircle of triangle APQ opposite to the vertex A, it follows that the point C is the center of the excircle of triangle APQ opposite to the vertex A. Thus, this point C lies on the external angle bisector of the angle APQ. Hence, $\measuredangle CPQ=90^{\circ}-\frac{\measuredangle APQ}{2}$. Similarly, $\measuredangle CQP=90^{\circ}-\frac{\measuredangle AQP}{2}$. Hence, by the sum of the angles in triangle PCQ, we have $\measuredangle PCQ=180^{\circ}-\measuredangle CPQ-\measuredangle CQP=180^{\circ}-\left(90^{\circ}-\frac{\measuredangle APQ}{2}\right)-\left(90^{\circ}-\frac{\measuredangle AQP}{2}\right)$ $=\frac{\measuredangle APQ+\measuredangle AQP}{2}$. But since the triangle PAQ is right-angled at A (this is because the quadrilateral ABCD is a square), we have < APQ + < AQP = 90°, and thus $\measuredangle PCQ=\frac{90^{\circ}}{2}=45^{\circ}$. darij

Find the point R in line AB but not in segment AB, which satisfies that BR=DQ. Then we can show triangle CDQ and triangle CBR are congruent triangles very easily by using S.A.S. SO we can say that angle PCQ is equal to the half of right angle RCQ.

Hi, How did you know that PCQ was exactly half of RCQ? I followed you with extending AB and marking point R, and the triangle congruence, but I'm lost after that. Thanks

You can then prove that CPQ and CPR are congruent triangles because PQ=2-AQ-AP=(1-AQ)+(1-AP)=DQ+BP=BR+BP=PR.

using trigo it;s easy, but how do we prove that if $ (tgx+1)(tgy+1)=2$ then $ x+y=45$?

Quote: An easy extension. Let $ ABCD$ be a quadrilateral for which $ AB=AD$, $ BC=CD$ and $ B=D=90^{\circ}$, $ m(\widehat{PAQ})=2\phi$ (a right deltoid). Let $ P\in (AB)$, $ Q\in (AD)$ be two points so that $ PQ=PB+QD$. Show that $ m(\widehat{PCQ})=90^{\circ}-\phi$. Proof. Observe that the semiperimeter of the triangle $ APQ$ is equal to $ AB$, $ CB\perp AP$ and the point $ C$ belongs to the bisector of the angle $ \widehat{PAQ}$. Therefore, the point $ C$ is the $ A$- exincenter of the triangle $ APQ$. Generally, $ m(\widehat{PCQ})=\frac{1}{2}\cdot \left(m(\widehat{APQ})+m(\widehat{AQP})\right)$, i.e. $ \boxed{m(\widehat{PCQ})=90^{\circ}-\phi}$.

Tigerli wrote: You can then prove that CPQ and CPR are congruent triangles because PQ=2-AQ-AP=(1-AQ)+(1-AP)=DQ+BP=BR+BP=PR. The 2 triangles are congruent as per s.s.s. criterion. Best regards, sunken rock