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Incircle of Delta ABC touches the sides BC, CA and AB at D, E and F, respectively. Let r_(1) be the radius of incircle of DeltaBDF. Then prove that r_(1) = (1)/(2) ((s-b) sin B)/((1+ sin.(B)/(2))) |
Answer» Solution : `ANGLE DIF = pi -B` Now, `FD = 2FP = 2r sin ((pi -B)/(2)) = 2r cos.(B)/(2)` Also, `BD = BF = s - b` Now, in -radius of `DeltaBDF` is `r_(1)` `:. r_(1) ("AREA of "DeltaBDF)/("Semiperimeter of " DeltaBDF)` `= ((1)/(2) (s-b)^(2) sinB)/((1)/(2) (2(s-b) + 2r cos.(B)/(2)))` `= ((s-b) sin B)/(2(1 + tan.(B)/(2) cos.(B)/(2))) "" ( :' r = (s-b) tan.(B)/(2))` `= ((s-b) sin B)/(2(1+ sin.(B)/(2))) ( :' r = (s-b) tan.(B)/(2))` |
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