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Q.

In a triangle ABC, bisector of angle C meets the side AB at D and circumcentre at E. The maximum value of CD. DE is equal to 

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a

a2/4

b

b2/4

c

c2/4

d

(a+b)2/4

answer is C.

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Detailed Solution

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CD.DE=AD.DB

AD+DB=c

Question Image

Since  A.M.  G.M. 

AD+DB2ADDB

or       c24ADDB

ADDBc2/4

and the required value is c2/4.

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