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In a δ Abc, the Internal Bisectors of ∠B and ∠C Meet at P and the External Bisectors of ∠B and ∠C Meet at Q, Prove that ∠Bpc + ∠Bqc = 180°. - Mathematics

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In a Δ ABC, the internal bisectors of ∠B and ∠C meet at P and the external bisectors of ∠B and ∠C meet at Q, Prove that ∠BPC + ∠BQC = 180°.

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In the given problem, BP and CP are the internal bisectors of ∠B and ∠C respectively. Also, BQ and CQ are the external bisectors of ∠B and ∠C respectively. Here, we need to prove:

∠BPC +∠BQC = 180°

We know that if the bisectors of angles ∠ABC and ∠ACB of ΔABC meet at a point O then` ∠BOC = 90° + 1/2 ∠A`.

Thus, in ΔABC

 `∠BPC = 90° + 1/2 ∠A.`            ……(1)

Also, using the theorem, “if the sides AB and AC of a ΔABC are produced, and the external bisectors of ∠B and ∠Cmeet at O, then `∠BOC = 90° + 1/2 ∠A.`  .

Thus, ΔABC

\[\angle BQC = 90^\circ - \frac{1}{2}\angle A . . . . . . \left( 2 \right)\]

Adding (1) and (2), we get

`∠BOC +∠BQC  = 90° + 1/2 ∠A + 90° -1/2 ∠A`

∠BQC + ∠BQC = 180°

Thus, `∠BQC + ∠BQC = 180°

Hence proved.

  Is there an error in this question or solution?
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RD Sharma Mathematics for Class 9
Chapter 11 Triangle and its Angles
Exercise 11.2 | Q 8 | Page 21
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