Nickolas Taylor

2022-07-14

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...

My attempt : used base division method to find the area of ASN but I get extra variables which is tough...

Maggie Bowman

Beginner2022-07-15Added 14 answers

$[.]$ 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 $[BPM]=[SPB]$

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

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 $[BPM]=[SPB]$

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

Keenan Santos

Beginner2022-07-16Added 1 answers

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

For $\mathrm{\u25b3}AMR$,

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

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

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

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

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

$=[ABD]-[ASN]$

For $\mathrm{\u25b3}AMR$,

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

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

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

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

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

$=[ABD]-[ASN]$

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