The points \(A\), \(B\), \(D\), and \(E\) lie on the circumference of a circle with centre \(C\), as shown.
If \(\angle BCD=72\degree\) and \(CD=DE\), then determine the measure of \(\angle BAE\).
We draw radii from \(C\) to points \(A\) and \(E\) on the circumference, and join \(B\) to \(D\). Since \(CA\), \(CB\), \(CD\), and \(CE\) are all radii, \(CA = CB = CD = CE\).
We’re given that \(CD=DE\). Since \(CD=CE\), we have \(CD=CE=DE\), and thus \(\triangle CDE\) is equilateral. It follows that \(\angle ECD = \angle CED = \angle CDE = 60\degree\).
Let \(\angle CDB = x^{\circ}\),
\(\angle CBA = y^{\circ}\), and \(\angle CAE = z^{\circ}\).
Since \(CB=CD\), \(\triangle CBD\) is isosceles. Therefore,
\(\angle CBD=\angle
CDB=x\degree\).
Since \(CA=CB\), \(\triangle CAB\) is isosceles. Therefore,
\(\angle CAB=\angle
CBA=y\degree\).
Since \(CE=CA\), \(\triangle CEA\) is isosceles. Therefore,
\(\angle CEA=\angle CAE=z\degree\).
Since the angles in a triangle sum to \(180\degree\), from \(\triangle CBD\) we have \(x\degree +x\degree +72\degree=180\degree\). Thus, \(2x\degree=108\degree\) and \(x=54\).
Since \(ABDE\) is a quadrilateral and the sum of the interior angles of a quadrilateral is equal to \(360\degree\), we have \[\begin{aligned} \angle BAE +\angle ABD +\angle BDE +\angle DEA&=360\degree\\ (y\degree +z\degree)+(y\degree +x\degree)+(x\degree +60\degree)+(60\degree +z\degree)&=360\degree\\ 2x+2y+2z&=240\\ x+y+z&=120 \\ 54+y+z&=120 \\ y+z&=66 \end{aligned}\]
Since \(\angle BAE =(y+z)\degree\), then \(\angle BAE=66\degree\).