\(\triangle PQR\) is right-angled at \(Q\). Point \(X\) lies on \(PQ\), point \(Z\) lies on \(QR\), and point \(Y\) lies on \(PR\) such that \(PX=PY\) and \(RZ=RY\).
Determine the measure of \(\angle XYZ\).
Solution 1
Let \(\angle RYZ = a\degree\) and \(\angle PYX = b\degree\).
Since \(RY=RZ\), \(\triangle RYZ\) is isosceles and \(\angle RZY=\angle RYZ = a\degree\).
Since \(PY=PX\), \(\triangle PYX\) is isosceles and \(\angle PXY=\angle PYX = b\degree\).
The angles in a triangle sum to \(180\degree\). Therefore, in \(\triangle RYZ\), \[\begin{aligned}
\angle YRZ + \angle RYZ + \angle RZY &= 180\degree\\
\angle YRZ + a\degree + a\degree &= 180\degree\\
\angle YRZ &= 180\degree - 2a\degree
\end{aligned}\] In \(\triangle
PQR\), \[\begin{aligned}
\angle RPQ +\angle PQR + \angle QRP&= 180\degree\\
\angle RPQ +90\degree+(180\degree-2a\degree)&=180\degree\\
\angle RPQ &= 2a\degree - 90\degree
\end{aligned}\] In \(\triangle
PYX\), \[\begin{aligned}
\angle PYX + \angle PXY + \angle YPX &= 180\degree\\
b\degree + b\degree+\angle YPX &= 180\degree
\end{aligned}\] Since \(\angle YPX =
\angle RPQ\) (same angle), we have \[\begin{aligned}
b\degree + b\degree+(2a\degree-90\degree) &= 180\degree\\
2b\degree &= 270\degree- 2a\degree\\
b\degree&=135\degree -a\degree
\end{aligned}\]
Now, \(PYR\) forms a straight line, so \(\angle PYR=180\degree\). That is, \[\begin{aligned} \angle PYX + \angle XYZ + \angle RYZ&=180\degree\\ b\degree+\angle XYZ + a\degree &= 180\degree\\ (135\degree -a\degree)+\angle XYZ + a\degree &= 180\degree\\ \angle XYZ&=180\degree-135\degree = 45\degree \end{aligned}\] Therefore, \(\angle XYZ = 45\degree\). Note that in solving for \(\angle XYZ\) it is was not necessary to determine either \(a\degree\) or \(b\degree\).
Solution 2
In \(\triangle PQR\), let \(\angle RPQ = m\degree\) and \(\angle PRQ = n\degree\). The angles in a triangle sum to \(180\degree\). Therefore, in \(\triangle PQR\), \[\begin{aligned} \angle RPQ + \angle PQR + \angle PRQ &= 180\degree\\ m\degree + 90\degree+n\degree &= 180\degree\\ m\degree +n\degree &= 90\degree \end{aligned}\] Therefore, \(m+n=90\).
Since \(PY=PX\), \(\triangle PYX\) is isosceles and so \(\angle PXY=\angle PYX\).
Also, in \(\triangle PYX\) \[\begin{aligned} \angle PYX + \angle PXY + \angle YPX &= 180\degree\\ \angle PYX + \angle PYX+m\degree &= 180\degree\\ 2\angle PYX &= 180\degree- m\degree \end{aligned}\] Therefore, \(\angle PXY = \angle PYX= 90\degree - \left(\frac{m}{2}\right)\degree\).
Similarly, since \(RY=RZ\), \(\triangle RYZ\) is isosceles, and therefore \(\angle RYZ=\angle RZY\). Also, in \(\triangle RYZ\), \[\begin{aligned} \angle RYZ + \angle RZY + \angle YRZ &= 180\degree\\ \angle RYZ + \angle RYZ+n\degree &= 180\degree\\ 2\angle RYZ &= 180\degree- n\degree \end{aligned}\] Therefore, \(\angle RYZ= \angle RZY= 90\degree - \left(\frac{n}{2}\right)\degree\).
Since \(\angle PYR=180\degree\), we have \[\begin{aligned} \angle PYX + \angle XYZ + \angle RYZ&=180\degree\\ \left( 90\degree - \left(\frac{m}{2}\right)\degree\right)+\angle XYZ + \left(90\degree - \left(\frac{n}{2}\right)\degree\right) &= 180\degree\\ 180\degree - \left(\frac{m}{2}\right)\degree - \left(\frac{n}{2}\right)\degree+ \angle XYZ&=180\degree\\ \angle XYZ&= \left(\frac{m}{2}\right)\degree + \left(\frac{n}{2}\right)\degree\\ &= \left(\frac{m+n}{2}\right)\degree \end{aligned}\] Therefore, since \(m+n=90\), we have \(\angle XYZ= \left(\frac{90}{2}\right)\degree = 45\degree\).
Therefore, \(\angle XYZ = 45\degree\). Note that in solving for \(\angle XYZ\) it is was not necessary to determine either \(m\degree\) or \(n\degree\).