UPSC 2013 CDS 2 Math Paper

Draw DG || BF.
In ΔADG, E is the midpoint of AD and EF || GD
∴AF = FG.....(1) (Midpoint Theorem)
In ΔBCF D is the midpoint of BC and DG || BF.
∴ CG = FG.....(2) (Midpoint Theorem)
From (1) and (2), we have
AF=FG=CG=

AC

3

=

2

3

AC