# ABCD is a parallelogram, P is any point on AC. Through P, MN is drawn parallel to BA cutting BC in

ABCD is a parallelogram, P is any point on AC. Through P, MN is drawn parallel to BA cutting BC in M and AD in N. SR is drawn parallel to BC cutting BA in S and CD in R. Show that [ASN]+[AMR]=[ABD] (where $\left[.\right]$ denotes the area of the rectilinear figure).

My attempt : used base division method to find the area of ASN but I get extra variables which is tough...
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Maggie Bowman
$\left[.\right]$ represents area
Draw $DP$ and $BP$ and using the properties of parallelogram (diagonal bisects the area)
$\left[PMR\right]=\left[CMR\right]$
$\left[ASN\right]=\left[PSN\right]$
$\left[DPR\right]=\left[APR\right]$ (same base, equal height) and $\left[DPR\right]=\left[NPD\right]$
$\left[BPM\right]=\left[APM\right]$ (same base, equal height) and $\left[BPM\right]=\left[SPB\right]$
$\left[PMR\right]+\left[ASN\right]+\left[APR\right]=\left[APM\right]=\frac{\left[ABCD\right]}{2}=\left[ABD\right]$
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Keenan Santos
Note $\left[ASN\right]=\frac{1}{2}\left[ASPN\right]$, $\left[ABD\right]=\frac{1}{2}\left[ABCD\right]$.
For $\mathrm{△}AMR$,
$\left[AMR\right]=\left[ABCD\right]-\left[ABM\right]-\left[CRM\right]-\left[ARD\right]$
$=\left[ABCD\right]-\frac{1}{2}\left[ABMN\right]-\frac{1}{2}\left[CRPM\right]-\frac{1}{2}\left[ASRD\right]$
$=\left[ABCD\right]-\frac{1}{2}\left(\left[ABMN\right]+\left[CRPM\right]+\left[ASRD\right]\right)$
$=\left[ABCD\right]-\frac{1}{2}\left(\left[ABCD\right]+\left[ASPN\right]\right)$
$=\frac{1}{2}\left[ABCD\right]-\frac{1}{2}\left[ASPN\right]$
$=\left[ABD\right]-\left[ASN\right]$