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Question
In the given figure, PB is the bisector of ABC and ABC =ACB. Prove that:
a. BC x AP = PC x AB
b. AB:AC = BP: BC
Solution
a. Construction: Draw CE || BP and produce AB to E.
Proof: BP || EC
∠PBC = ∠BCE ...(Alternate angles)
∠ABP = ∠AEC ...(Corresponding angles)
Also, ∠ABP = ∠PBC
⇒ ∠BCE = ∠BEC
So, BE = BC
In ΔAEC,
⇒ `"AP"/"PC" = "AB"/"BE"`
⇒ BC x AP = PC x AB.
b. Note: It is not possible to prove this part due to inadequate data.
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