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Asked by Balaji Last Modified
Answer 
Tasneem
Let us consider a triangle ABC in which CD ⊥ AB.
It is given that
cos A = cos B
… (1)
We have to prove ∠A = ∠B. To prove this, let us extend AC to P such that BC = CP.
From equation (1), we get
(through construction we have BC=CP) .............(2)
By using the converse of B.P.T,
CD||BP
⇒∠ACD = ∠CPB (Corresponding angles) … (3)
And, ∠BCD = ∠CBP (Alternate interior angles) … (4)
By construction, we have BC = CP.
∴ ∠CBP = ∠CPB (Angle opposite to equal sides of a triangle) … (5)
From equations (3), (4), and (5), we obtain
∠ACD = ∠BCD … (6)
In ΔCAD and ΔCBD,
∠ACD = ∠BCD [Using equation (6)]
∠CDA = ∠CDB [Both 90°]
Therefore, the remaining angles should be equal.
∴∠CAD = ∠CBD
⇒ ∠A = ∠B
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