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Question
In ΔABC prove that `(b + c - a) tan "A"/(2) = (c + a - b)tan "B"/(2) = (a + b - c)tan "C"/(2)`.
Solution
`(b + c - a) tan "A"/(2)`
= `(a + b + c - 2a). sqrt(((s - b)(s - c))/(s(s - a)`
= `(2s - 2a).sqrt(((s - b)(s - c))/(s(s - a)`
= `2sqrt(((s - a)(s - b)(s - c))/s` ....(1)
`(c + a - b) tan "B"/(2)`
= `(a + b + c - 2b). sqrt(((s - a)(s - c))/(s(s - b)`
= `(2s - 2b).sqrt(((s - a)(s - c))/(s(s - b)`
= `2sqrt(((s - a)(s - b)(s - c))/s` ...(2)
`(a + b - c) tan "C"/(2)`
= `(a + b + c - 2c). sqrt(((s - a)(s - b))/(s(s - c)`
= `(2s - 2c).sqrt(((s - a)(s - b))/(s(s - c)`
= `2sqrt(((s - a)(s - b)(s - c))/s` ...(3)
From (1), (2) an (3), we get
`(b + c - a)tan "A"/(2) = (c + a - b)tan "B"/(2) = (a + b - c)tan "C"/(2)`.
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