Question

PLS HELP ME GUYS
ABCD is a parallelogram in which angle DAB = 80 degrees. Bisector of angle A and angle B meets CD at P. Prove that :
1. AD=DP
2. CP=CB
3. DC=2AD​

Answer 1

Step-by-step explanation:

∠ C = 80°

Now, ∠D = ∠ B = 100°

So, AD = DP .

∠ BPA = 90° as ∠ PBA = CBP

∠BPC = 50° = ∠PBC

So, CP = CB

Also, DC = 2AD

DC = DP + PC

= AD + CB = AD + AD

= 2AD

Answer 2

Given: angle DAB = 80, Bisector of angle A and angle B meets CD at P.

To find: Prove that :

                1. AD=DP

                2. CP=CB

                3. DC=2AD

Solution:

• Now we have given that AP and BP is the angle bisector, so

            ang C = 80°

            ang DAP = 40

• Now, ∠D = ∠ B = 100°

• So, In triangle DAP,

             ang A + ang P + ang D = 180  .............(by angle sum property of a triangle)

• So, AD = DP . ..............(i)

• Now,

              ang BPA = 90° ............... (ang PBA =  ang CBP )

       also,

              ang BPC = 50° = ang PBC

• So, CP = CB    ....................(ii)

• So,

                 DC = DP + PC

                AD + CB = AD + AD

                = 2AD..............(iii)

Answer:

• So we have proved that AD=DP, CP=CB and DC=2AD​ in i, ii and iii.