Page 18 - MATINF Nr. 13-14

18 E.A.J. Garc´ıa

hence
cot(α/2) 1
.
= p
(s − b)(s − c) (s − a)(s − b)(s − c)(s − d)
» cot(α/2) 1
By Brahmagupta’s formula, ∆ = (s − a)(s − b)(s − c)(s − d), so = . Cyclic
(s − b)(s − c) ∆
relabeling of a, b, c, d and α, β, γ, δ yields the other three equalities.

Reduction to the classical law of cotangents (triangle case d = 0).

Let d = 0. Then D coalesces with A and ABCD degenerates to the triangle ABC with
a + b + c
semiperimeter s = and area ∆ (see Figure 2).
2

Figure 2. Since d = 0, D coalesces with A.

Taking α as the angle formed by AB and the limiting tangent at A, the tangent–chord
theorem [2] and the limiting relation α + γ = π give α = π − γ, hence cot(α/2) = tan(γ/2).

From the generalized theorem, with d = 0,

cot(β/2) 1 cot(γ/2) 1
= , = . (1)
(s − c)s ∆ (s − a)s ∆

2
Let r be the inradius of 4ABC. Since ∆ = rs and (by Heron) ∆ = s(s − a)(s − b)(s − c),
we have
(s − a)(s − b)(s − c)
2
2 2
r s = s(s − a)(s − b)(s − c) =⇒ r = . (2)
s
Using (1) and ∆ = rs,
cot(β/2) 1 cot(γ/2) 1
= , = . (3)
s − c r s − a r

0
0
To obtain the relation at A, set α = ∠BAC. Since α = π − (β + γ),
α 0 Å β + γ ã tan(β/2) + tan(γ/2)
cot = tan = .
2 2 1 − tan(β/2) tan(γ/2)