In the following figure, ABC is an equilateral triangle and P is any point in AC; prove that :

(i) BP > PA

(ii) BP > PC

(i) Since, ABC is an equilateral triangle.

∴ ∠A = ∠B = ∠C = 60°.

In △ ABP,

∠ABP = ∠B - ∠PBC

∴ ∠ABP < ∠B

∴ ∠ABP < ∠A (Since, ∠B = ∠A)

∴ PA < BP or BP > PA [If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.]

Hence, proved that BP > PA.

(ii) In △ BPC,

∠PBC = ∠B - ∠ABP

∴ ∠PBC < ∠B

∴ ∠PBC < ∠C (Since, ∠B = ∠C)

∴ PC < BP or BP > PC [If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.]

Hence, proved that BP > PC.