Q21 of 95 Page 6

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

Given: DEFG is a square and ∠BAC = 90°

To Prove: DE2 = BD × EC.


In AGF and DBG


∠GAF = ∠BDG [each 90°]


∠AGF = ∠DBG


[corresponding angles because GF|| BC and AB is the transversal]


∴AFG ~ DBG [by AA Similarity Criterion] …(1)


In AGF and EFC


∠GAF = ∠CEF [each 90°]


∠AFG = ∠ECF


[corresponding angles because GF|| BC and AC is the transversal]


∴AGF ~EFC [by AA Similarity Criterion] …(2)


From equation (1) and (2), we have


DBG ~ EFC


Since, the triangle is similar. Hence corresponding sides are proportional



[∵DEFG is a square]


⇒DE2 = BD × EC


Hence Proved


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(i) ΔAMC ~ ΔPNR


(ii)


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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.

 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

 23

In the given figure, ∠ABC = 90° and BD ⊥ AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.