Question

ABCD is a parallelogram having AB > AD. From AB cut off AE equal to AD. Show that DE bisects angle ADC . P is a point in AB produced such that PD bisects angle APC . Prove that PC = AB .​

Answer 1

Step-by-step explanation:

R.E.F image

∴DC∥AB or DC∥AP

∴DC=

2

1

AP... by mid-point theorem

DC=AB=BP

CB∥AD or CB∥RA

∴CB=

2

1

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=

2

1

AP...(i)

CB=

2

1

AR...(ii)

Comparing (i) and (ii)

⇒

2

1

AP=

2

1

AR

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.

solution

Answer 2

Step-by-step explanation:

please see the above photo for the answer ✌❤❣