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ABCD is a parallelogram in which A = 60o. If bisectors of A and B meets at P. Prove that AD = DP, PC = BC, DC = 2AD

Asked by Topperlearning User 4th June 2014, 1:23 PM
Answered by Expert
Answer:

(i) ABCD is a parallelogram in which A = 60°.

AP and PB are the bisector of A and B respectively.

PAB = PAD = 30° (AP bisects A)

AB || CD and AP is the transversal.

PAB = APD = 30° (Alternate angles)

In APD,

PAD = APD = 30°

PD = AD ...(1) (Equal sides have equal angles opposite to them in a trangle)

(ii) AD || BC and AB is the transversal.

A + B = 180° (Sum of adjacent interior angles is 180°)

60° + B = 180°

B = 120°

PBA = PBC = 60° (PB is bisector B)

PBA = BPC = 60° (Alternate angles)

In PBC,

BPC = PBC = 60°

BC = PC ...(2) (Equal sides have equal angles opposite to them)

(iii) CD = DP + PC

CD = AD + BC [Using (1) and (2)]

CD = AD + AD [AD = BC (Opposite sides of parallelogram are equal)]

CD = 2AD

Answered by Expert 4th June 2014, 3:23 PM
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