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Question
In triangle ABC; ∠A = 60o, ∠C = 40o, and the bisector of angle ABC meets side AC at point P. Show that BP = CP.
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Solution
In ΔABC,
∠A = 60°
∠C = 40°
∴ ∠B = 180° - 60° - 40°
⇒ ∠B = 80°
Now,
BP is the bisector of ∠ABC.
∴ ∠PBC = `"∠ABC"/2`
⇒ ∠PBC = 40°
In ΔPBC,
∠PBC = ∠PCB = 40°
∴ BP = CP ...[Sides opp. to equal angles are equal.]
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