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Problem of the Week
Problem E and Solution
Find Another Angle


The points \(A\), \(B\), \(D\), and \(E\) lie on the circumference of a circle with centre \(C\), as shown.

Points A, B, D, and E are placed, in order,
around the circle with A and D on opposite sides of the circle and
sector BCD with an angle of 72 degrees. Line segments AB, BC, CD, DE,
and EA form a five-sided figure inside the circle.

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, \(\angle CEA=\angle CAE=z\degree\).

The interior of cyclic quadrilateral ABDE is divided into
four isosceles triangles with the centre of the circle, C, as a common

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