Triangle Relationship?

Question

In triangle ABC, suppose that angle C is twice angle A. Show that

ab = c² - a²

A, B, and C are the angles and a, b, and c are the lengths of the corresponding sides.

Answer 1

Let x be the angle A
a² = b² + c² - 2bc cos x ----------(1)
c² = a²+ b² - 2ab cos 2x ----------(2)

(a/sin x) = (c/sin 2x)
(a/sinx) = (c/2sin x cos x)
2a cos x = c

c² = a² + b² - 2ab cos 2x
c² = a² + b² - 2ab (2cos² x - 1)
c² = a² + b² - 2a cos x (2b cos x) + 2ab
c² = a² + b² - 2bc cos x + 2ab
c² = a² + b² - (b² + c² - a²) + 2ab
c² = a² - c² + a² + 2ab
2c² - 2a² = 2ab
c² - a² = ab

Finally I can figure out the solution...This question is quite interesting!!!

Answer 2

hmm maybe it works like this
apply generalized Pytaghora theorem to find cos A and cosC in terms of sides. Apply formula for cos(2x) in terms of cos(x)
no...dead end.