Question

If O be any point inside the triangle ABC , prove that angle BOC is greater than angle BAC

Answer 1

see the pic

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Answer 2

Hence Proved.

Step-by-step explanation:

Given,

$O$ be any point inside the $\triangle ABC$ then prove that $\angle BOC > \angle BAC$

Let,

Draw a perpendicular line $AD$ with $BC$ through $O$ as shown in the figure.

We know,

The exterior angle of a triangle is equal to the sum of the two opposite interiors angle.

Exterior angle of a triangle$=$Sum of the two opposite interiors angle

∴ $\angle BOD=\angle ABO+\angle OAB$

⇒ $\angle BOD> \angle OAB$__1

And $\angle COD=\angle OCA+\angle OAC$

⇒$\angle COD> \angle OAC$__2

Add both sides of equation-1 & 2,

  $\angle BOD+\angle COD> \angle OAB+\angle OAC$

∴ $\angle BOC > \angle BAC$    ( where $\angle BOD+\angle COD=\angle BOC$ and $\angle OAB+\angle OAC=\angle BAC$ )

Hence Proved.