The height BD is drawn in triangle ABC (then D lies on the side AC). It turned out that AB = 2CD

The height BD is drawn in triangle ABC (then D lies on the side AC). It turned out that AB = 2CD and CB = 2AD. Find the angles of triangle ABC.

The height BD is drawn in triangle ABC (then D lies on the side of AC). It turned out that AB = 2CD and CB = 2AD. Find the angles of triangle ABC.

Because the height is lowered to the side of the AC at a right angle of 90 degrees, then it divided our triangle into two right-angled triangles with a common leg BD. Let’s designate side AB as 2a, DC as a, BC as 2b, AD as b. By the Pythagorean theorem, we find that BD ^ 2 = 4a ^ 2 – b ^ 2 and BD ^ 2 = 4b ^ 2 – a ^ 2, whence we find 4a ^ 2 – b ^ 2 = 4b ^ 2 – a ^ 2;

whence 5a ^ 2 = 5b ^ 2, a = b, AC = 2a = 2b, which means we have an equilateral triangle ABC where each angle is 60 degrees.

Answer: Angle ABC = Angle BCA = Angle CAB = 60 degrees.