Please solve q 57 ?

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1 Answer
CW
May 11, 2018

CD+CE=S/2

Explanation:

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Let O and r be the center and the radius of the incircle, as shown in the figure.
Given AB+BC+CA=S
Let |ABC| denote area of DeltaABC,
=> |ABC|=1/2*r*(AB+BC+CA)=1/2rS
given DE divides DeltaABC into two equal areas,
=> |CDE|=1/2|ABC|=1/4rS ----- Eq(1)
|CDE|=1/2*r*(CD+CE) ----- Eq(2)
As Eq(2)=Eq(1),
=> 1/2*r*(CD+CE)=1/4rS
=> CD+CE=S/2

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