Let us assume the triangle ABC in which CD⊥AB

Give that the angles A and B are acute angles, such that

Cos (A) = cos (B)

As per the angles taken, the cos ratio is written as

AD/AC = BD/BC

Now, interchange the terms, we get

AD/BD = AC/BC

Let take a constant value

AD/BD = AC/BC = k

Now consider the equation as

AD = k BD …(1)

AC = k BC …(2)

By applying Pythagoras theorem in △CAD and △CBD we get,

CD² = BC² – BD² … (3)

CD² =AC² −AD² ….(4)

From the equations (3) and (4) we get,

AC²−AD² = BC²−BD²

Now substitute the equations (1) and (2) in (3) and (4)

K²(BC²−BD²)=(BC²−BD²) k²=1

Putting this value in equation, we obtain

AC = BC

∠A=∠B (Angles opposite to equal side are equal-isosceles triangle)