In figure, ∆ABC, ∠B = 2 ∠C. If bisector of the ∠ABC meets AC in P, prove that
a. Cb/ba= cp/oa
b. AB x BC = BP x CA
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Step-by-step explanation:
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Answered by
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Step-by-step explanation:
ABC
∠ABC=2∠ACB ( given)
Let ∠ACB=x
BP is bisector of ∠ABC
∴∠ABP=∠PBC=x
By angle bisector theorem
the bisector of an divides the side opposite to it in ratio of other two sides
Hence,
BC
AB
=
PA
CP
∴
BA
CB
=
PA
CP
Consider △ABC and △APB
∠ABC=∠APB (exterior angle properly)
given that, ∠BCP=∠ABP
∴ABC≈△APB (AA criteria)
∴
BP
AB
=
CB
CA
(corresponding sides of similar triangle are propositional)
∴AB×BC=BP×CA
Hence proved
solution
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