On the BC side of the parallelogram ABCD, point e is taken, that CE = CD, the angle ADE

On the BC side of the parallelogram ABCD, point e is taken, that CE = CD, the angle ADE is 55 degrees. Find the angle C of the parallelogram.

According to the theorem about the angles formed by two parallel lines and a secant:
1. If two parallel lines are intersected by a secant, the angles lying crosswise are equal.
2. If two parallel straight lines are crossed by a secant, then the corresponding angles are equal.
3. If two parallel straight lines are intersected by a secant, then the sum of one-sided angles is 180 °,
∠DEC = ∠ADE = 55 °.
In order to find the angle C of the parallelogram, consider the triangle ΔECD.
If EC = CD, then the triangle is isosceles, and in an isosceles triangle the angles at the base are equal. Means:
∠DEC = ∠CDE = 55 °.
Since in a triangle the sum of all its angles is 180 °, then:
∠DCE = 180 ° – ∠DEC – ∠CDE;
∠DCE = 180 ° – 55 ° – 55 ° = 60 °.
Answer: degree measure ∠С = 60 °.