Answer :

By the property that says, if a perpendicular is drawn from the vertex of a right triangle to the hypotenuse then the triangles on both the sides of the perpendicular are similar to the whole triangle and also to each other.

We can conclude by the property in ∆BDC,

∆CQD ∼ ∆DQB

(a). To Prove: DQ^{2} = DP × QC

Proof: As already proved, ∆CQD ∼ ∆DQB

We can write the ratios as,

By cross-multiplication, we get

DQ^{2} = QB × QC …(i)

Now since, quadrilateral PDQB forms a rectangle as all angles are 90° in PDQB.

⇒ DP = QB & PB = DQ

And thus replacing QB by DP in equation (i), we get

DQ^{2} = DP × QC

Hence, proved.

(b). To Prove: DP^{2} = DQ × AP

Prof: Similarly using same property, we get

∆APD ∼ ∆DPB

We can write the ratios as,

By cross-multiplication, we get

DP^{2} = PB × AP

⇒ DP^{2} = DQ × AP [∵ PB = DQ]

Hence, proved.

Rate this question :