In triangle ABC, angle A = 90, BD- bisector, angle ADB = 50/1) Find the angles of triangle BDC.

In triangle ABC, angle A = 90, BD- bisector, angle ADB = 50/1) Find the angles of triangle BDC. 2) Compare BD and CD

In a right-angled triangle AED, we determine the value of the angle ABD.

Angle ABD = (180 – 90 – 50) = 40.

Since BD, by condition, is the bisector of the angle ABC, then the angle CBD = ABD = 40, and therefore the angle ABC = 2 * 40 = 80.

Then in a right-angled triangle ABC the angle ACB = (180 – 90 – 80) = 10.

In the triangle ВDС we define the value of the angle СDВ. Angle СDВ = (180 – 40 – 10) = 130.

The segment BD is located opposite the corner 10, and the segment CD is located opposite the angle 40, therefore, BD <CD.

Answer: The angles of the triangle ВСD are equal to 10, 40, 130. ВD <СD.