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In A B C , as semicircle is inscribed, ...

In ` A B C ,` as semicircle is inscribed, which lies on the side `cdot` If `x` is the lengthof the angle bisector through angle `C ,` then prove that the radius of the semicircle is `xsin(C/2)dot`

Text Solution

Verified by Experts


From the figure
`((1)/(2)) ra + ((1)/(2)) rb = ((1)/(2)) ab sin C`
or `r(a + b) = 2 Delta`
or `r = (2Delta)/(a + b) = (ab sin C)/(a + b)`
Also `x = (2ab)/(a + b) cos. (C)/(2)` [length of angle bisector]
From Eq. (i), `r = (2 xx (1//2) ab sin C)/(a + b)`
`= (2 ab sin.(C)/(2) cos. (C)/(2))/(a + b)`
`= (2 ab cos.(C)/(2))/(a + b) sin. (C)/(2)`
`= x "sin" (C)/(2)`
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