Question

Prove that
(S-a)tanA/2=(s-b)tan B/2=(s-c)tanC/2

Answer 1

Before we solve this problem, we must know some formulae:

a/sinA = b/sinB = c/sinC ..... (1)

cosA = (b² + c² - a²)/(2bc)

cosB = (c² + a² - b²)/(2ca)

cosC = (a² + b² - c²)/(2ab)

Proof:

From (1), we write

a/sinA = b/sinB = c/sinC = k (≠ 0)

Then sinA = a/k, sinB = b/k, sinC = c/k

1st term = (s - a) tan(A/2)

= {(a + b + c)/2 - a} * sinA/(1 + cosA)

= (b + c - a)/2 * [(a/k) / {1 + (b² + c² - a²)/(2bc)}]

= (b + c - a)/2 * 2abc/k * 1/(b² + 2bc + c² - a²)

= abc (b + c - a)/k * 1/{(b + c)² - a²}

= abc (b + c - a)/k * 1/{(b + c + a)(b + c - a)}

= abc/{k (a + b + c)}

i.e, (s - a) tan(A/2) = abc/{k (a + b + c)}

Similarly using tan(B/2) = sinB/(1 + cosB) and tan(C/2) = sinC/(1 + cosC) from the last two terms given, we can obtain

(s - b) tan(B/2) = abc/{k (a + b + c)} and

(s - c) tan(C/2) = abc/{k (a + b + c)}

Therefore,

(s - a) tan(A/2) = (s - b) tan(B/2) = (s - c) tan(C/2)

Hence proved.