In triangle ABC BD, the bisector of angle B, angle A = 90, AD = √5, BC = 2√5. find the area of the triangle BDC.

Since ВD, by condition, is the bisector of the angle, then it divides the sides of the AC into segments proportional to the adjacent sides.
AD / AВ = СD / ВС.
√5 / AB = СD / 2 * √5.
AB * СD = √5 * 2 * √5 = 10.
Segment AB is the height of the ВDС triangle to the base of the СD, then:
the area of the ВСD triangle is equal to: Svsd = СD * AB / 2 = 10/2 = 5 cm2.
Answer: The area of the ВСD triangle is 5 cm2.