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Answer:
Given △ABC in which BD is the bisector of ∠B and a line PQ||AC meets AB,BC and BD at P,Q and R respectively.
Proof (i)
In △BQP, BR is the bisector of ∠B.
∴
BP
BQ
=
PR
QR
⇒ BQ.PR=BP.QR
⇒ PR.BQ=QR.BP [Hence proved]
(ii) In △ABC, we have
PQ∣∣AC [Given]
⇒
AP
AB
=
CQ
CB
[By Thale's Theorem]
⇒ AB×CQ=BC.AP [Hence proved]
Step-by-step explanation:
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