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

$$ABCD$$ 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

• Hint : Use triangle = 1/2 parallelogram. – cosmo5 Oct 8 '20 at 9:16
• Can you pls provide me the solution i have tried so many times but everytime i got stuck – anonymous Kumar Oct 8 '20 at 10:09

2 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 $$[BPM]=[SPB]$$

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

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

For $$\triangle AMR$$,

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

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

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

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

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

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