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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: ∠APB is a right angle.
Solution
∠DAP =∠PAB
∠CBP = ∠PBA
∠DAB + ∠CBA = 180° ...(adjacent angles of || gm are supplementary)
Multipying by `(1)/(2)`
`(1)/(2)∠"DAB" + (1)/(1)∠"CBA" = (1)/(2) xx 180°`
∠PAB + ∠PBA = 90°
In ΔAPB,
∠PAB + ∠PBA + ∠APB = 180°
90° + ∠APB = 180°
∠APB = 90°
Therefore, ∠APB is a right angle.
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