Question

P is the midpoint of side BC of parallelogram ABCD coma such that angle 1 = angle 2. prove that AD=2CD​

Answer 1

Answer:

AD=2CD

Step-by-step explanation:

Given : ABCD is a parallelogram.

P is the mid point of BC and ∠1 = ∠2 so,

∠BAP = ∠DAP

we have To prove that AD = 2 CD

Proof : Given, ∠BAP = ∠DAP

∴ ∠2 = ∠BAP = 1/2 ∠BAD  ....................(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,

∠3 + ∠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

BC = 2AB   so,

AD = 2CD

Answer 2

Answer:

Step-by-step explanation:

GIVEN: ABCD is a parallelogram

           ∠1 = ∠2

           P is the midpoint of BC

TO PROVE: AD = 2 CD

Proof:  Since ABCD is a a parallelogram,

AD ║ BC                 [opposite sides of a ║gm are parallel and equal]

Ap is transversal since it intersects both AD and BC. Hence

∠1 = ∠3                   [alternate interior angles pair] -------------------------(1)

Also, ∠1 = ∠2         [given in the question]      --------------------------------(2)

From (1) and (2) we conclude, ∠2= ∠3

In ΔABP, since ∠2 = ∠3

⇒ AB = BP                        [sides opposite to equal angles in a Δ are equal]

Now, in ║gm ABCD, AD = BC                [opposite sides of a ║gm are equal]

⇒ AD = BP  +  PC

⇒AD = BP   +   BP                                                     [P is the midpoint of BC]

⇒AD = 2 BP

⇒AD = 2AB                               [BP=AB  proved ]

⇒AD = 2CD                            [CD = AD; opposite sides of a ║gm are equal]