Question

In a triangle ABC BD is internal angle bisector of angle B and CD is external angle bisector of angle C. A line DQ is drawn parallel to BC, which intersects AC and AB at point P and Q respectively , if PC=8cm , PQ=2 cm then find BQ=?

Answer 1

Given : In a triangle ABC BD is internal angle bisector of angle B and CD is external angle bisector of angle C. A line DQ is drawn parallel to BC, which intersects AC and AB at point P and Q respectively ,  PC=8cm , PQ=2 cm

To Find : BQ

Solution:

BD is internal angle bisector of angle B

=> ∠CBD = ∠ABD   = ∠B/2

QD || BC  & BD transversal

=> ∠QDB = ∠CDB  alternate interior angles

=>  ∠QDB = ∠ABD

=> ∠QDB = ∠QBD

=> BQ = DQ  sides opposite to equal angle

CD is external angle bisector of angle C

=> ∠PCD = (1/2) external angle C

∠PDC =  (1/2) external angle C ( alternate interior angle  ) as QD || BC  & CD transversal

=> PC = PD

BQ = DQ

=> BQ = PD + PQ

=> BQ = PC + PQ

=> BQ = 8 + 2

=> BQ = 10 cm

BQ = 10 cm

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