# 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 $$\angle BAE=66\degree$$.