#
Problem
of the Week

Problem
E and Solution

Find
Another Angle

## Problem

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 \(\angle BAE\).

## Solution

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 \(\angle BAE=66\degree\).