Question

ABCD is a parallelogram with AB greater than AD. From AB cut off AB
to AD. Show that DE bisects the angle ADC. Find a point P in AB prod
> Prove that is
B4 meet Ay at Hand Cy at K and prove that H
39
such that PD bisects the angle APC​

Answer 1

Answer:

BT bisects ∠ABC and meets AD at T. A straight line through C and parallel to BT meets AB produced at P and AD produced at R. Prove that ΔRAP is isosceles and the sum of two equal sides of the ΔRAP is equal to the perimeter of the parallelogram ABCD.

December 27, 2019avatar

Satyajit Champ

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ANSWER

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

Answer:

see the above photo for answer ❣❤✌