Triangle Relationship?

Question

Given triangle ABC with corresponding sides of length a, b, and c. In triangle ABC, Show that c² = (b² - a²)²/(a² + b²).

Answer 1

take sine on both sides

sin(B) = sin(90 + A)

sin(B) = cos(A)

Now
sin(C) = sin(180 -(A+B)) = sin(A+B)

but B = 90 + A

sin(C) = sin(A + 90 + A)

sin(C) = sin(90 + 2A)

sin(C) = cos(2A)

using sine rule

a/sinA = b/sinB = c/sinC

but sin(B) = cos(A) and sin(C) = cos(2A)

a/sinA = b/cosA = c/cos(2A) = k , some constant

a = k sinA --------------(1)

b = k cosA ----------------(2)

c = k cos(2A)--------------(3)

from (1) and (2)

a^2 + b^2 = k^2 -------------(4)

from (3)

c = k (2cos^2(A) - 1)

squaring

c^2 = k^2[2 (cos^2(A) -1)^2

=>c^2 = k^2[4 cos^4(A) + 1 - 4cos^2(A)]

from (2) cos(A) = b/k

=>c^2 = k^2[4 (b/k)^4 + 1 - 4(b/k)^2]

multiply with k^2/k^2

c^2 = [4 b^4/k^2 + k^2 - 4b^2]

from (4), k^2 = a^2 + b^2

c^2 = [(4 b^4/(a^2+b^2)) + a^2 + b^2 - 4b^2]

c^2 = [4b^4 + (a^2 + b^2)^2 - 4b^2(a^2+b^2)]/(a^2 +b^2)

c^2 = [4b^4 + a^4 + b^4 + 2a^2b^2 - 4b^2a^2 - 4b^4]/(a^2+b^2)

c^2 = [b^4 + a^4 - 2b^2a^2]/(a^2 + b^2)

c^2 = [(b^2 - a^2)^2]/(a^2 + b^2)

Answer 2

drop a line from any point to the side opposite and give it a letter like "d". Then do the next = "e" and then finally "f". The segments of the line are subdivided into a1 and a2, b1 and b2 and c1 and c2. Now you have a series of right sided triangles for which you can use the Pythagorean rule x^2 + y^" = z^2 where z is the side opposite the right angle. This gives you a series of simultaneous equations which then need to be solved.