Question

In a right triangle ABC, right-angled at B, D is a point on hypotenuse such that $BD\sim AC$. If $DP\sim AB$ and $DQ\sim BC$ then prove that

(a) $D{Q}^2=DP.QC$

(b) $D{P}^2=DQ.AP$

Answer 1

We know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangles on the both sides of the perpendicular are similar to the whole triangle and also to each other.
(a) Now using the same property in In △BDC, we get

△CQD ∼ △DQB


Now. since all the angles in quadrilateral BQDP are right angles.
Hence, BQDP is a rectangle.
So, QB = DP and DQ = PB

∴

(b)

Similarly, △APD ∼ △DPB

[∵DQ=PB]