Question

4. In a ABC, AB >AC. The bisectors of B and C meet at
P. Prove that BP > CP.​

Answer 1

Answer:

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Step-by-step explanation:

Given A △ABC in which the bisectors of ∠B and ∠C meet the sides AC and AB at D and E respectively.

To prove AB=AC

Construction Join DE

Proof In △ABC, BD is the bisector of ∠B.

∴  BCAB=DCAD...........(i)

In △ABC, CE is the bisector of ∠C.

∴  BCAC=BEAE.......(ii)

Now, DE∣∣BC

⇒   BEAE=DCAD            [By Thale's Theorem]......(iii)

From (iii), we find the RHS of (i) and (ii) are equal. Therefore, their LHS are also equal i.e.

      BCAB=BCAC

⇒   AB=AC

Hence,  △ABC is isosceles.