Question

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

Explanation:

Step1: Identify the shapes

The figure consists of a square \(ABCD\) and an equilateral triangle \(\triangle ABE\). Since \(AB = BC=CD = AD\) (square - all sides equal) and \(AB = AE=BE\) (equilateral triangle - all sides equal), we know that \(AE = AD\) and \(BE=BC\).

Step2: Analyze angles in square and equilateral - triangle

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

Step3: Calculate \(\angle DAE\) and \(\angle CBE\)

\(\angle DAE=\angle BAD-\angle BAE\). Since \(\angle BAD = 90^{\circ}\) and \(\angle BAE=60^{\circ}\), then \(\angle DAE = 90 - 60=30^{\circ}\). Similarly, \(\angle CBE = 30^{\circ}\).

Step4: Consider isosceles - triangles \(\triangle ADE\) and \(\triangle BCE\)

In \(\triangle ADE\), \(AD = AE\), so \(\triangle ADE\) is isosceles. Let \(\angle AED=\angle ADE\). Using the angle - sum property of a triangle (\(\angle DAE+\angle AED+\angle ADE = 180^{\circ}\)), and since \(\angle AED=\angle ADE\), we have \(2\angle AED=180 - 30\), so \(\angle AED=\frac{180 - 30}{2}=75^{\circ}\). Similarly, in \(\triangle BCE\), \(\angle BEC = 75^{\circ}\).

Step5: Calculate \(\angle AEC\)

\(\angle AEC=\angle AEB-(\angle BEC+\angle AED)\). \(\angle AEB = 60^{\circ}\), \(\angle BEC = 75^{\circ}\), \(\angle AED = 75^{\circ}\). So \(\angle AEC=360-(90 + 90+60)=120^{\circ}\).

Answer:

\(120\)