Q20 of 95 Page 6

In the adjoining figure, the diagonal BD of a parallelogram ABCD intersects the segment AE at the point F, where E is any point on the side BC. Show that DF x FE = BF x FA.

Given: ABCD is a parallelogram

To Prove: DF x FE = BF x FA


In AFD and BFE


∠1 = ∠2 (alternate angles)


∠3 = ∠4 (vertically opposite angles)


∴ AFD ~BFE (by AA similarity criterion)


So,


(corresponding sides of similar triangle are proportional)



⇒ DF x FE = BF x FA


Hence Proved


More from this chapter

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18

In the given figure, OA .OB = OC.OD, show that: ∠A = ∠C and ∠B = ∠D.

 19

In the given figure, CM and RN are respectively the medians of ΔABC and ΔPQR. If ΔABC ~ ΔPQR, prove that:

(i) ΔAMC ~ ΔPNR


(ii)


(iii) ΔCMB ~ ΔRNQ


 21

In the given figure, DEFG is a square and ∠BAC is a right angle. Show that DE2= BD x EC.

 22

In the given figure, ABD is a right angled triangle being right angled at A and AD ⊥ BC. Show that:


(i) AB2= BC.BD


(ii) AC2 = BC. DC


(iii) AB. AC. = BC. AD