Question

ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively intersecting at P, Q and R. Prove that the perimeter of ΔPQR is double the perimeter of ΔABC.

Answer 1

Perimeter of ΔPQR = 2(Perimeter of ΔABC)

Step-by-step explanation:

Given data:

ABC is a triangle.

ABCQ and ARBC are parallelograms.

∴ BC = AQ and BC = AR

⇒ AQ = AR

⇒ A is the mid-point of QR.

Similarly, B and C are the midpoints of PR and PQ respectively.

By definition of midpoint theorem,

$A B=\frac{1}{2} P Q$

PQ = 2AB ------ (1)

$B C=\frac{1}{2} Q R$

QR = 2BC ------ (2)

$C A=\frac{1}{2} P R$

PR = 2CA ------ (3)

Add (1), (2) and (3), we get

PQ + QR + PR = 2AB + 2BC + 2CA

PQ + QR + PR = 2(AB + BC + CA)

Perimeter of ΔPQR = 2(Perimeter of ΔABC)

Hence perimeter of ΔPQR is double the perimeter of ΔABC.

To learn more...

1. ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 m, prove that the area of ∆APQ is one-sixteenth of the area of ∆ABC.

https://brainly.in/question/6200073

2. A circle is touching the side BC of triangle ABC at P and touching AB and AC producced at Q and R respectively.Prove that AO=1/2 (perimeter ABC).

https://brainly.in/question/1025330

Answer 2

Given : In

Δ

A

B

C

,

Through A, B and C, lines are drawn parallel to BC, CA and AB respectively meeting at P, Q and R.

To prove : Perimeter of

Δ

P

Q

R

=

2

×

p

e

r

i

m

e

t

e

r

o

f

Δ

A

B

C

Proof :

∵

PQ|| BC and QR||AB

∴

ABCQ is a || gm

∴

BC = AQ

Similarly , BCAP is a ||gm

∴

BC = AP ....(i)

∴

AQ = AP = BL

→

PQ = 2BC

Similarly, we can prove that

QR = 2AB and PR = 2AC

Now perimeter of

Δ

P

Q

R

.

= PQ+QR+PR = 2AB+2BC+2AC

=2(AB+BC+AC)

= 2 perimeter of

Δ

A

B

C

Hence proved.