In the isosceles triangle DEP, the bisector PM of angle P is drawn at the base of DP, ∡PME = 78 °.

In the isosceles triangle DEP, the bisector PM of angle P is drawn at the base of DP, ∡PME = 78 °. Determine the angles of this triangle.

Since the triangle DEP is isosceles, the angle EDP = EPD.

Let the value of the angle ЕDР = X0. Since PM is the bisector of the angle DPE, the angle DPM = (X / 2) 0.

The PME angle is the external angle of the PDM triangle, the value of which is equal to the sum of the internal angles that are not adjacent to it.

Angle PME = MDP + DPM = X + X / 2 = 3 * X / 2.

78 = 3 * X / 2.

3 * X = 156.

X = EDP = EPD = 156/3 = 52.

Then the angle DEP = (180 – 52 – 52) = 76.

Answer: The angles of the triangle DEP are 52, 52, 76.