In parallelogram ABCD, angle A is 65 degrees. Sections CC1 and CC2 are its heights. Find the value of the angle С1СС2.

The angle C1AC2 is equal to the angle C2DC as the corresponding angles at the intersection of parallel straight lines AC2 and BC of the secant AC1, then the angle C1BC = 65, and the angle C1CB = 180 – 90 – 65 = 25.

The angle C1AC2 is equal to the angle C2DC as the corresponding angles at the intersection of parallel straight lines AC1 and DC of the secant AC2, then the angle C2DC = 65, and the angle C2CD = 180 – 90 – 65 = 25.

Angle ВСD = BAD = 65, as opposed angles of a parallelogram, then the angle С1СС2 = ВСD + С1СВ + С2СD = 65 + 25 + 25 = 115.

Answer: The angle С1СС2 is 115.




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