Question

19. In a right triangle ABC, right-angled at B,
D is a point on hypotenuse such that
BD 1 AC. If DP I AB and DQ 1 BC then
prove that
(a) DQ2 = DP.QC (b) DP2 = DQ: AP.​

Answer 1

Answer:

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Answer 2

Answer:

we can conclude by the property in ∆BDC

∆CQD ~ ∆ DQB

( a ) . To prove : DQ2 ,=DP × QC

proof: As already proved ,∆CQD ~ ∆DQB

we can write the ratios as ,

CQ ÷ DQ = DQ ÷ QB

by cross multiplication , we got

DQ2 = QB × QC ...... ( 1 )

Now since , quadrilateral PDQB forms a

rectangle as all angles are 90 ° in PDQB

DP = QB & PB = DQ

As thus replacing QB by DP in equation ...1

we get

DQ2 = DO × AP

Hence , Proved