In a square ABCD, E is a point inside the square such that DEC is an equilateral triangle. If E is joined to vertices A and B of the square, what is the degree measure of AEB?
Show answer & explanation
Correct answer: B
Explanation
The correct answer is option 2 i.e. 150°.
Here, AD = DC = CB, and DC = DE = EC
So, AD = DC = CB = DE = EC
∠ADC = ∠BCD = 90°
∠ADE = ∠ADC - ∠EDC = 90° - 60° = 30°
∠BCE = ∠BCD - ∠ECD = 90° - 60° = 30°
In triangle ADE, AD = DE. So, it's an isosceles triangle.
Then, ∠DAE = ∠DEA
In triangle ADE,
∠ADE + ∠DAE + ∠DEA = 180°
⇒ 30° + 2∠DEA = 180°
⇒ 2∠DEA = 150°
⇒ ∠DEA = 75°
In triangle BCE, CB = CE
So, it's an isosceles triangle, and ∠CBE = ∠CEB = 75°
The sum of the angles around point E is 360°.
So,
∠AEB + ∠DEA + ∠DEC + ∠CEB = 360°
⇒ ∠AEB + 75° + 60° + 75° = 360°
⇒ ∠AEB = 360° - 210° = 150°
Therefore, the degree measure of angle AEB is 150°.