Question

abcd is parallegram if the bisectos dp and cp of angles d and c meet at p on side ab then show that p is the mid point of side ab​

Answer 1

Since ABCD is a parallelogram, then AB || CD  and  AD || BC.

Also, AD = BC  ;  AB = CD   [ opposite sides of parallelogram are equal]    

…..(1)

Since PD bisects ∠∠ADC, then

∠ADP = ∠PDC∠ADP = ∠PDC      …(2)

Since AB || CD and PD is a transversal, then

∠APD = ∠PDC     [alternate interior angles]∠APD = ∠PDC     [alternate interior angles]     …..(3)

from (2) and (3), we get

∠ADP = ∠APD

In ΔAPD,

we have

∠ADP = ∠APD⇒AP = AD  [sides opposite to equal angles in a triangle are equal]  …(4)

Since ​ PC bisects ∠∠BCD, then

∠BCP = ∠PCD∠BCP = ∠PCD      ….(5)

Since AB || CD and PC is a transversal, then

∠BPC = ∠PCD   [alternate interior angles]∠BPC = ∠PCD   [alternate interior angles]   ….(6)

from (5) and (6), we have

∠BCP = ∠BPC

In ΔPBC,

we have

∠BCP = ∠BPC⇒PB = BC   [sides opposite to equal angles in a triangle are equal]  …(7)

∠BCP = ∠BPC

    AD = BC

⇒⇒AP = PB    [using (4) and (7)]

⇒⇒P is the mid-point of AB.

Answer 2

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