Maths-

General

Easy

Question

# ABC is a triangle in which AB = AC. P is any point in its interior such

that. Prove that AP bisects

Hint:

### find relation between angle B and angle C using AB = AC. Now ,

Find relation between ∠PBC and ∠PCB Also relation between PB and PC.

Now prove and are congruent so , corresponding angles are equal.

## The correct answer is: Hence proved

### In ABC,

AB = AC (given)

∠B = ∠C(angles opposite to equal sides are equal)

here, ∠ ABP = ∠ ACP (given)

From diagram we get,

∠B = ∠ ABP + ∠ PBC —-Eq1

∠C = ∠ ACP + ∠ PCB—-Eq2

Subtracting Eq1-Eq2 we get,

∠B - ∠C = ∠ ABP + ∠ PBC - ∠ ACP - ∠ PCB

Substitute ∠ ABP = ∠ ACP and ∠ B = ∠ C,

We get ∠ PBC = ∠ PCB

In PBC, ∠ PBC = ∠PCB

Then , PB = PC (side opposite to equal angles are equal)

We get PB = PC(side )

∠ABP = ∠ ACP (given)(Angle)

AB = AC (given)(side)

By SAS rule,

By congruence , ∠ BAP = ∠ CAP then , ∠ BAC is bisected by AP

Hence proved

From diagram we get,

Subtracting Eq1-Eq2 we get,

By congruence , ∠ BAP = ∠ CAP then , ∠ BAC is bisected by AP

Hence proved

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