UPSC 2014 CDS II Math Paper

Here, $BC \parallel DE$, $AB \parallel EF$ and $AC \parallel DF$Remember: If both pairs of opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram.Therefore, the quadrilaterals BCAD,= ABFC and ABCE are parallelograms.In the parallelogram BCAD,$BC = AD$....(1)Remember:Opposite sides of a In the parallelogram ABCE,$BC = AE$ … (2) parallelogram are equal.Adding (a) and (b), we have $2BC = AD + AE = DE$Similarly, $DF = 2AC$ and $EF = 2AB$Hence, statement 1 is correct.In $\triangle ABD$ and $\triangle ABC$$AD = BC$ (Opposite sides of a parallelogram are equal)$BD = AC$ (Opposite sides of a parallelogram are equal)$AB = AB$ (Common)$\text{therefore} \triangle ABD = \triangle ABC$ (SSS congruence rule)$\Rightarrow \text{ar}(\triangle ABD) = \text{ar}(\triangle ABC)$Similarly, $\text{ar}(\triangle ABC) = \text{ar}(\triangle ACE)$ and $\text{ar}(\triangle ABC) = \text{ar}(\triangle BCF)$ $\text{therefore} \text{ar}(\Delta ABC) = \frac{1}{4} \text{ar}(\Delta DEF)$$\Rightarrow \text{ar}(\triangle DEF) = 4 \times \text{ar}(\triangle ABC)$Hence, statement 2 is correct