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Question
ABCD is a parallelogram. The bisector of ∠BAD meets DC at P, and AD is half of AB.
Prove that: BP bisects ∠ABC.
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Solution
Since PC = BC ...(AD is half of AB and BC = AD and DC = AB)
∠CPB = ∠CBP
But ∠CPB = ∠PBA ...(alternate angles ∵ DC || AB)
⇒ ∠CBP = ∠PBA
Therefore, BP bisects ∠ABC.
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