In a right triangle ABC, right angled at B, D is a point on hypotenuse such that BD⊥AC.

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

$\triangle \mathrm{CQD} \sim \triangle \mathrm{DQB}$

$\frac{\mathrm{CQ}}{\mathrm{DQ}}=\frac{\mathrm{DQ}}{\mathrm{QB}}$

$\Rightarrow \mathrm{DQ}^{2}=\mathrm{QB} . \mathrm{CQ}$

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

$\therefore \mathrm{DQ}^{2}=\mathrm{DP} . \mathrm{CQ}$

(b)

Similarly, △APD ∼ △DPB

$\frac{\mathrm{AP}}{\mathrm{DP}}=\frac{\mathrm{PD}}{\mathrm{PB}}$

$\Rightarrow \mathrm{DP}^{2}=\mathrm{AP} . \mathrm{PB}$

$\Rightarrow \mathrm{DP}^{2}=\mathrm{AP} . \mathrm{DQ} \quad[\because \mathrm{DQ}=\mathrm{PB}]$