If ∠ A and ∠ B are acute angles such that Cos A = Cos B, then show that ∠ A = ∠ B

Let us consider a triangle ABC in which CD ⊥ AB

It is given that

Cos A = Cos B

We have to prove ∠A = ∠B. To prove this, let us extend AC to P such that BC = CP

From equation (i), we obtain

  (By construction BC = CP)     (ii)

By using the converse of B.P.T,

CD||BP

∠ACD = ∠CPB (Corresponding angles)     (iii)

And,

∠BCD = ∠CBP (Alternate interior angles)     (iv)

By construction, we have BC = CP

∠CBP = ∠CPB (Angle opposite to equal sides of a triangle)     (v)

From equations (iii), (iv), and (v), we obtain

∠ACD = ∠BCD     (vi)

In ΔCAD and ΔCBD,

∠ACD = ∠BCD      [Using equation (vi)]

∠CDA = ∠CDB        (Both 90°)

Therefore, the remaining angles should be equal

∠CAD = ∠CBD

∠A = ∠B