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
(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)
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)
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
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