Question

ABCD is a parallelogram. BT bisects ABC and meets AD and T. A straight li  ne through



C and parallel to BT meets AB produced at P and AD produced at R. Prove that RAP  is

equal to the perimeter of the parallelogram ABCD.​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

R.E.F image

∴DC∥AB or DC∥AP

∴DC= ½ AP... by mid-point theorem

DC=AB=BP

CB∥AD or CB∥RA

∴CB= ½ RA... by mid-point theorem

CB=DA=DR

Let ABCD be a square (∵ square is also a ∥ gram)

∴DC=CB=BA=DA

Here, DC=CB

and, DC=½

AP...(i)

CB= ½ AR...(ii)

Comparing (i) and (ii)

⇒ ½AP= ½AR

Multiplying 2 both sides

⇒AP=AR

∴ΔRAP is an isosceles triangle

Now,

To show :- AP+AR= Perimeter of ∥ gram ABCD

We know that,

Perimeter of ∥ gram ABCD=AB+BC+CD+DA

but =AB+BC=AP and AD+DC=AR (from ab)

∴AP+AR= Perimeter of ABCD

Hence proved✅.