Question

The bisectors of base angles of a triangle cannot enclose a right angle in any case.

Answer 1

In ΔABC sum of all angles of a triangle is 180°

∠A + ∠B + ∠C = 180°

Divide both sides by 2, we get

\(\frac{1}{2}\)∠A + \(\frac{1}{2}\)∠B + \(\frac{1}{2}\)∠C = 180°

\(\frac{1}{2}\)∠A + ∠OBC + ∠OCB = 90°[Therefore, OB, OC bisects ∠B and ∠C]

∠OBC + ∠OCB = 90° - \(\frac{1}{2}\)∠A

Now, in ΔBOC

∠BOC + ∠OBC + ∠OCB = 180°

∠BOC + 90° - \(\frac{1}{2}\)∠A = 180°

∠BOC = 90° - \(\frac{1}{2}\)∠A

Hence, bisector open base angle cannot enclose right angle.