In the adjoining figure, P is mid-point of the side BC of a parallelogram ABCD such that ∠BAP = ∠DAP. Prove that AD = 2CD.
Answers
Given : ABCD is a parallelogram. P is the mid point of BC and ∠BAP = ∠DAP
To prove : AD = 2 CD
Proof : Given, ∠BAP = ∠DAP
∴ ∠1 = ∠BAP = 1/2 ∠A ...(1)
ABCD is a parallelogram,
∴ AD || BC (Opposite sides of the parallelogram are equal)
∠A + ∠B = 180° (Sum of adjacent interior angles is 180°)
∴ ∠B = 180° – ∠A ...(2)
In ΔABP,
∠1 + ∠2 + ∠B = 180° (Angle sum property)
=> 1/2∠A + ∠2 + 180 - ∠A = 180 [Using equations (1) and (2)]
=> ∠A - 1/2 ∠A = 0
=> ∠A = 1/2 ∠A ...(3)
From (1) and (2), we have
∠1 = ∠2
In ΔABP,
∠1 = ∠2
∴ BP = AB (In a triangle, equal angles have equal sides opposite to them)
=> 1/2 BC = AB (P is the midpoint on BC)
=> BC = 2AB
⇒ AD = 2CD (Opposite sides of the parallelogram are equal)
Hence, proved.
Given : P is the mid-point of the side BC of a parallelogram ABCD
∠BAP = ∠DAP
To Find : Prove that AD = 2CD.
Solution:
Draw a line PQ || AB Q is point on AD
=> ∠QAP = ∠DAP
∠BAP = ∠DAP
=> ∠BAP = ∠QAP
∠QAP = ∠BPA ( alternate angle as AQ || PB ∵ AD || BC)
∠BAP = ∠BPA
=> AB = BP
P is mid point of BC
=> BC = 2BP
BC = AD
=> AD = 2BP
=> AD = 2AB
AB = CD
=> AD = 2CD
QED
Hence proved
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