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

Question

$ABCD$ 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 $[.]$ 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...

Help

Answer 1

$[.]$ represents area

Draw $DP$ and $BP$ and using the properties of parallelogram (diagonal bisects the area)

$[PMR]=[CMR]$

$[ASN]=[PSN]$

$[DPR]=[APR]$ (same base, equal height) and $[DPR]=[NPD]$

$[BPM]=[APM]$ (same base, equal height) and $[BPM]=[SPB]$

$[PMR]+[ASN]+[APR]=[APM]=\frac {[ABCD]}{2}=[ABD]$

Answer 2

Note $[ASN] = \frac{1}{2}[ASPN]$, $ [ABD] = \frac{1}{2}[ABCD]$.

For $\triangle AMR$,

$$ [AMR] = [ABCD] - [ABM] - [CRM] - [ARD] $$

$$ = [ABCD] - \frac{1}{2}[ABMN] - \frac{1}{2}[CRPM] - \frac{1}{2}[ASRD] $$

$$ = [ABCD] - \dfrac{1}{2}([ABMN] + [CRPM] + [ASRD]) $$

$$ = [ABCD] - \dfrac{1}{2}([ABCD] + [ASPN]) $$

$$ = \frac{1}{2}[ABCD] - \frac{1}{2}[ASPN] $$

$$ = [ABD] - [ASN] $$