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The bisectors of ∠ B and ∠ C of an isosceles △ ABC with AB = AC intersect each other at a point O. Show that the exterior angle adjacent to ∠ ABC is equal to ∠ BOC.

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Considering the △ ABC

It is given that AB = AC

So we get

∠ ABC = ∠ ACB

Dividing by 2 both sides

½ ∠ ABC = ½ ∠ ACB

So we get

∠ OBC = ∠ OCB …… (1)

By using the angle sum property in △ BOC

∠ BOC + ∠ OBC + ∠ OCB = 180o

Substituting equation (1)

∠ BOC + 2 ∠ OBC = 180o

So we get

∠ BOC + ∠ ABC = 180o

From the figure we know that ∠ ABC and ∠ ABP form a linear pair of angles so we get

∠ ABC + ∠ ABP = 180o

∠ ABC = 180o – ∠ ABP

By substituting the value in the above equation we get

∠ BOC + (180o – ∠ ABP) = 180o

On further calculation

∠ BOC + 180o – ∠ ABP = 180o

By subtraction

∠ BOC – ∠ ABP = 180o – 180o

∠ BOC – ∠ ABP = 0

∠ BOC = ∠ ABP

Therefore, it is proved that the exterior angle adjacent to ∠ ABC is equal to ∠ BOC.

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