Question

In triangle ABC, ∠ACB=90∘. D and E are points on AB such that AD=DE=EB. If there exists θ such that CD=5cosθ and CE=5sinθ, what is AB^2?

Answer 1

i got it using co-ordinate geometry

Answer 2

\( AB^2 = x^2+y^2 \\ AD =DE =EB = \sqrt {x^2+y^2}/3 \\ CD^2 = (x/3)^2+(2y/3)^2 = 25\cos^2 \theta \\ CE^2=(2x/3)^2+(y/3)^2=25\sin^2 \theta\)
Squaring and adding the 2 above equations, you'll get x^2+y^2