ABC is a triangle, right-angled at C and Ac = √3 BC. Prove that ABC = 60°.
Given: △ABC is right angled at C and AC = √3BC.
To prove: ∠ABC = 60°.
Let D be the midpoint of AB. Join CD.
Now, AB2 = BC2 + AC2 = BC2 + (√3BC)2 = 4BC2
Therefore AB = 2BC.
Now, BD = 1/2 AB = 1/2(2BC) = BC.
But, D being the midpoint of hypotenuse AB, it is equidistant from all the
Therefore CD = BD = DA or CD = 1/2 AB = BC.
Thus, BC = 80 = CD,
i.e., △BCD is a equilateral triangle.
Hence, ∠ABC = 60°.