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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