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Quadrilaterals — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsQuadrilaterals5 Marks
Question
In the adjoining figure, $\triangle\text{ABC}$ is a triangle and through $A, B, C$, lines are drawn, parallel respectively to $BC, CA$ and $AB$, intersecting at $P, Q$ and $R$. Prove that the perimeter of $\triangle\text{PQR}$ is double the perimeter of $\triangle\text{ABC}.$
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Answer

Given: A $\triangle\text{ABC},$ in which through points $A, B$ and $C$, lines $QR, QP$ and $RP$ have been draw parrallel to $BC, AC$ and $AB$ of $\triangle\text{ABC}$ respectively
.
To Prove: Perimeter of $\triangle\text{PQR}=2(\text{Perimeter of} \triangle\text{ABC})$
Proof: Since $AR\ ||\ BC$ and $AB\ ||\ RC$ [Given] So, $ABCR$ is a parallelogram.
 Therefore $AR = BC ...(i)$ Also, $AQ\ ||\ BC$ and $QB\ ||\ AC$
So,$ AQBC$ is a parallelogram,
Therefore $QA = BC...(ii)$ Adding both side of $(i)$ and $(ii)$,
we get $AR + QA = BC + BC \Rightarrow QR = 2BC$
$\Rightarrow\text{BC}=\frac{\text{QR}}{2}$
$\therefore\text{BC}=\frac{1}{2}\text{QR}$
Similarly we can prove $\text{AB}=\frac{1}{2}\text{RP}$ and
$\text{AC}=\frac{1}{2}\text{PQ}$
So, Perimeter of $\triangle\text{PQR}$
 $= PQ + QR + RP = 2AC + 2BC + 2AB = 2(AC + BC + AB) =2$(Perimeter of $\triangle\text{ABC}$)

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