An equilateral triangle ABC surmounts a square BCDE such that side BC is common to both. Given this configuration, find the value of the expression:

$$\angle EAB + 3\angle AEB$$

Question

An equilateral triangle ABC surmounts a square BCDE such that side BC is common to both. Given this configuration, find the value of the expression:

$$\angle EAB + 3\angle AEB$$

Accepted Answer

**Solution:**

1. Since ABC is an equilateral triangle, all its sides are equal: $AB = BC = AC$.
2. BCDE is a square, so all sides are equal: $BC = CD = DE = EB$.
3. Because BC is common and equal in both, all sides of the combined figure are equal: $AB = BC = AC = CD = DE = EB$.

4. Triangle ABE is isosceles with $AB = EB$.

5. Angles in the equilateral triangle ABC are all $60^\circ$.
6. Angles in the square BCDE are all $90^\circ$.

7. Angle $\angle ABE$ is the sum of $\angle ABC$ (from triangle) and $\angle CBE$ (from square):
   $$\angle ABE = 60^\circ + 90^\circ = 150^\circ$$

8. In isosceles triangle ABE, base angles are equal:
   $$\angle BAE = \angle AEB$$

9. Sum of angles in triangle ABE:
   $$\angle BAE + \angle AEB + \angle ABE = 180^\circ$$
   $$2\angle AEB + 150^\circ = 180^\circ$$
   $$2\angle AEB = 30^\circ$$
   $$\angle AEB = 15^\circ$$
   Thus, $\angle BAE = 15^\circ$.

10. Since $\angle EAB = \angle BAE$, we have:
    $$\angle EAB = 15^\circ$$

11. Calculate the required expression:
    $$\angle EAB + 3\angle AEB = 15^\circ + 3 	imes 15^\circ = 15^\circ + 45^\circ = 60^\circ$$

**Answer:** $60^\circ$

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