Question

in the figure, ab = bc = cd = ad = ae = be. find the measure in degrees of ∠aec.

Explanation:

Step1: Identify the shapes

The figure consists of a square \(ABCD\) (since \(AB = BC=CD = AD\)) and an equilateral triangle \(\triangle ABE\) (since \(AB = AE=BE\)).

Step2: Find the angles of the square and equilateral - triangle

In a square, each interior angle is \(90^{\circ}\). In an equilateral triangle, each interior angle is \(60^{\circ}\).

Step3: Calculate \(\angle EAD\)

\(\angle BAD = 90^{\circ}\) and \(\angle BAE=60^{\circ}\), so \(\angle EAD=\angle BAD+\angle BAE = 90^{\circ}+60^{\circ}=150^{\circ}\).

Step4: Consider \(\triangle EAD\)

Since \(AD = AE\), \(\triangle EAD\) is isosceles. Let \(\angle AED=\angle ADE\). Using the angle - sum property of a triangle (\(\angle EAD+\angle AED+\angle ADE = 180^{\circ}\)), and since \(\angle AED=\angle ADE\), we have \(2\angle AED=180^{\circ}-\angle EAD\). Substituting \(\angle EAD = 150^{\circ}\), we get \(2\angle AED=180 - 150=30^{\circ}\), so \(\angle AED = 15^{\circ}\).

Step5: Calculate \(\angle AEC\)

Similarly, in \(\triangle EBC\), \(\angle EBC = 90^{\circ}+60^{\circ}=150^{\circ}\) and since \(BE = BC\), \(\angle BEC=\angle BCE = 15^{\circ}\).
\(\angle AEC=\angle AEB-\angle BEC-\angle AED\). Since \(\angle AEB = 60^{\circ}\), \(\angle AEC=60^{\circ}-15^{\circ}-15^{\circ}=30^{\circ}\).

Answer:

\(30\)