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=?
Answers
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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