Question

$L$ and $M$ are the mid-points of sides $AB$ and $DC$ respectively of parallelogram $ABCD$. Prove that segments $DL$ and $BM$ trisect diagonal $AC$.

Answer 1

The required figure is shown
From figure,
$BL=DM$ and $BL\parallel DM$ and $BLMD$ is a parallelogram, therefore $BM\parallel DL$
From triangle $ABY$
$L$ is the midpoint of $AB$ and $XY\parallel BY$, therefore $x$ is the midpoint of $AY$ .ie $AX=XY$ ..........(1)
Similarly for triangle $CDX$
$CY=XY$ ......(2)
From $\left(1\right)$ and $\left(2\right)$
$AX=XY=CY$ and $AC+XY+CY$
Hence proved