1.

Incircle of DeltaABC touches the sides BC, AC and AB at D, E and F, respectively. Then answer the following question The length of side EF is

Answer»

`r sin.(A)/(2)`
`2R sin.(A)/(2)`
`r cos.(A)/(2)`
`2r cos.(A)/(2)`

SOLUTION :
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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