Through the vertex B of triangle ABC, a straight line is drawn perpendicular to the height BD

Through the vertex B of triangle ABC, a straight line is drawn perpendicular to the height BD, calculate the angle ABD and the angle DBC if it is known that the angle BAC = 35 degrees and the angle BCA = 70 degrees.

Since the line MB is perpendicular to the height BD, and the height BD is perpendicular to the base of the AC, then MD is parallel to the AC.

Then the angle НBC = BCA = 70 as criss-crossing angles at the intersection of parallel straight lines AC and MВ secant BC.

Angle MBA = BAC = 35 as criss-crossing angles at the intersection of parallel straight lines AC and MB secant AB.

Then the angle ABD = MВD – MВA = 90 – 35 = 55.

Angle DВС = HBD – HBС = 90 – 70 = 20.

Answer: The ABD angle is 55, the DBS angle is 20.