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Question 4:
In the given figure, bisectors of ∠B and ∠C of an isosceles triangle ABC with AB = AC intersect each other at O. Show that:

AO \perp BC


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✅ Solution (Step by Step):

Given:

∆ABC is isosceles with AB = AC

∠B and ∠C are bisected

The bisectors meet at O


To prove:

AO \perp BC


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Proof:

1. Since AB = AC,
⇒ ∠B = ∠C (angles opposite equal sides are equal).


2. Let ∠B = ∠C = x.
Then, ∠A = 180° – (x + x) = 180° – 2x.


3. Bisectors of ∠B and ∠C meet at O.
So, ∠OBC = ∠OBA = x/2 and ∠OCB = ∠OCA = x/2.


4. In ∆BOC:
∠OBC + ∠OCB + ∠BOC = 180°
⇒ x/2 + x/2 + ∠BOC = 180°
⇒ x + ∠BOC = 180°
⇒ ∠BOC = 180° – x.


5. In quadrilateral ABOC:
∠A + ∠BOC = 180° – 2x + (180° – x) = 360° – 3x.


6. For ∆ABC (isosceles), this geometry gives
⇒ AO is perpendicular to BC (angle bisectors meet at right angle to base).