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 $Δ P Q R$ is double the perimeter of $Δ A B C$ .

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Solution

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 $Δ P Q R = 2 \times p e r i m e t e r o f Δ A B C$

Proof : $∵$ $P Q | | B C$ and $Q R | | A B$

$∴$ $A B C Q$ is a $| |$ gm

$∴$ $B C$ = $A Q$

Similarly , $B C A P$ is a $| |$ gm

$∴$ $B C$ = $A P$ ....(i)

$∴$ $A Q$ = $A P$ = $B L$

$\to$ $P Q$ = $2 B C$

Similarly, we can prove that

$Q R$ = $2 A B$ and $P R$ = $2 A C$

Now perimeter of $Δ P Q R .$

= $P Q + Q R + P R$ = $2 A B + 2 B C + 2 A C$

= $2 (A B + B C + A C)$

= $2$ perimeter of $Δ A B C$

Hence proved.

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