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Question

In ABC, B is a right angle, and BD is perpendicular to AC and DE is perpendicular to BC. Which of the following is true?


A

DC2×ED=EC2×AD

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B

ED2×DC=EC2×AD

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C

ED2×DC=AD2×EC

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D

DC2×ED=AD2×EC

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Solution

The correct option is B

ED2×DC=EC2×AD


In ABC and BDC

ABC=BDC=90

BCA=ECD (common in both the triangles)

Therefore, ABCBDC (by AA similarity criterion)

BCDC=ACBCBC2=AC×DC....(1)

Similarly, BDCDEC

Therefore, DCEC=BCDC
DC2=EC×BC

BC=DC2EC....(2)

On substituting (2) in (1), we get

DC4EC2 = AC×DC

DC3=EC2×AC

DC2×DC=EC2×(AD+DC)

(EC2+ED2)×DC=EC2(AD+DC)(DC2=EC2+ED2)

Therefore, ED2×DC=EC2×AD


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