Question
A $\triangle\text{ABC}$ is given. If lines are drawn through $A, B, C,$ parallel respectively to the sides $BC, CA$ and $AB$, forming $\triangle\text{PQR},$ as shown in the adjoining figure, show that $\text{BC}=\frac{1}{2}\text{QR}.$
✓
Answer
Given: A $\triangle\text{ABC}$ in which through points $A, B$ and $C$, line $QR, QP$ and $RP$ are drawn parallel to $BC, CA$ and $AB$.
To prove: $\text{BC}=\frac{1}{2}\text{QR}$
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 $AR + QA = BC + BC$
$\Rightarrow QR = 2BC$
$\Rightarrow\text{BC}=\frac{\text{QR}}{2}$
$\therefore\text{BC}=\frac{1}{2}\text{QR}$
To prove: $\text{BC}=\frac{1}{2}\text{QR}$
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 $AR + QA = BC + BC$
$\Rightarrow QR = 2BC$
$\Rightarrow\text{BC}=\frac{\text{QR}}{2}$
$\therefore\text{BC}=\frac{1}{2}\text{QR}$
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