Mathematics
ABC is an isosceles triangle in which AB = AC. P is any point in the interior of △ABC such that ∠ABP = ∠ACP. Prove that
(a) BP = CP
(b) AP bisects ∠BAC.
Triangles
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Answer
Below figure shows the isosceles triangle ABC with the points marked:
(a) Given,
AB = AC
∴ ∠B = ∠C ……..(i)
Given, ∠ABP = ∠ACP ……..(ii)
Subtracting (ii) from (i) we get,
∠B - ∠ABP = ∠C - ∠ACP
∠PBC = ∠PCB.
∴ BP = CP (As sides opposite to equal angles are equal)
Hence, proved that BP = CP.
(b) We know that,
BP = CP (Proved)
AB = AC (Given)
∠ABP = ∠ACP (Given)
Hence, △ABP ≅ △ACP by SAS axiom.
∠PAB = ∠PAC (Corresponding angles of congruent triangles are equal.)
Thus AP, bisects ∠BAC.
Hence, proved that AP bisects ∠BAC.
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