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In Fig. the incircle of △ABC, touches...

In Fig. the incircle of △ABC, touches the sides BC, CA and AB at D, E respectively. Show that : AF + BD + CE = AE + BF + CD = ½ (Perimeter of △ABC).

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The correct Answer is:
A

Points I, D, C, E are concylic. Therefore,
`angleEID = pi -C`
Also, I is circumcenter of `DeltaDEF`. Hence,
`angleDFE = (1)/(2) angleDIE = (pi -C)/(2)`
Similarly, `angleFDE = (pi -A)/(2)`
`angleFED = (pi -B)/(2)`
`:.` Area of `DeltaDEF = 2r^(2) sin (angleD) sin (angleE) sin (angleF)`
`= 2r^(2) "cos"(A)/(2) "cos"(B)/(2) "cos"(C)/(2)`
Also by sine rule, in `DeltaDEF`,
`(EF)/(sin angleD) = 2r`
or `EF = 2r cos.(A)/(2)`
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