Through $A, B$ and $C$ , lines $R Q, P R$ and $Q P$ have been drawn respectively, para llel sides $B C, C A$ and $A B$ of a $Δ A B C$ as shown in the figure. Show that $B C = \frac{1}{2} Q R$

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Solution

Given: Triangle $A B C$ and $P Q R$ in which $A B | | P Q, B C | | R Q$ and $C A | | P R$

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

Proof: Quadrilatral $R B C A$ is a parallelogram.

$(∵ R A | | B C a n d B R | | C A)$

$∴ R A = B C$ .......(1) ( $∵$ Opposite side of parallelogram)

Now, quadrilateral $B C Q A$ is a parallelogram.

$∴ A Q = B C$ ................(2)( $∵$ Opposite side of parallelogram)

Adding (1) and (2), we get,

$\Rightarrow R A + A Q = 2 B C$

$\Rightarrow Q R = 2 B C$

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

Hence proved.

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