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perpendiculars are drawn from the angles...

perpendiculars are drawn from the angles `A , Ba n dC` of an acute-angled triangle on the opposite sides, and produced to meet the circumscribing circle. If these produced parts are `alpha,beta,gamma` , respectively, then show that `a/alpha+b/beta+c/gamma=2(tanA+tanB+t a nC)dot`

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Let AD be the perpendicular from A on BC. When AD is produced, it meets the circumscribing circle at E. From question `DE = alpha`.

Since angle in the same segment are equal, we have
`angle AEB = angleACB = C`
and `angle AEC = angle ABC = B`
From the right-angled triangle BDE
`tan C = (BD)/(DE)`...(i)
From the right-angled triangle CDE,
`tan B = (CD)/(DE)`...(ii)
Adding Eqs. (i) and (ii), we get
`tan B + tan C = (BD + CD)/(DE) = (BC)/(DE) = (a)/(alpha)`....(iii)
Similarly, `tan C + tan A = (b)/(beta)`...(iv)
and `tan A + tan B = (c)/(gamma)`...(v)
Eqs. (iii), (iv), and (v), we get
`(a)/(alpha) + (b)/(beta) + (c)/(gamma) = 2 (tan A + tan B + tan C)`
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