Answer: To prove that △ABC is isosceles, we need to show that two of its sides are equal in length.
Given that AD is an external angle bisector of △ABC, we know that ∠CAD is half the measure of the opposite external angle ∠ABC. So, we have:
∠CAD = 1/2 * ∠ABC
We are also given that ∠CDE = 110° and ∠AEC = 30°.
Since AD is an external angle bisector, we can determine that ∠ADE = ∠CDE - ∠CAD. Plugging in the given values, we get:
∠ADE = 110° - 1/2 * ∠ABC
Similarly, we can find ∠AED using the fact that the angles in a triangle add up to 180°:
∠AED = 180° - ∠ADE - ∠AEC
Plugging in the given values, we have:
∠AED = 180° - (110° - 1/2 * ∠ABC) - 30°
Now, we can observe that ∠ADE = ∠AED since AD is an external angle bisector. So we can set the two expressions equal to each other:
110° - 1/2 * ∠ABC = 180° - (110° - 1/2 * ∠ABC) - 30°
Simplifying this equation gives:
110° - 1/2 * ∠ABC = 180° - 110° + 1/2 * ∠ABC - 30°
Combining like terms, we have:
1/2 * ∠ABC + 1/2 * ∠ABC = 110° + 30° - 180° + 110°
Simplifying further:
∠ABC = 60°
This shows that ∠ABC is equal to 60°. Since ∠ABC is one of the base angles of △ABC, and we have already proven that the two angles at the top of the triangle (formed by the external bisector) are congruent, we can conclude that △ABC is isosceles.