Question

Prove that, if a line parallel to a side of a triangle intersect the other sides in two distinct points, then the line divides those sides in proportion.

Answer 1

Given: In a $△PQR$, line $l\parallel$ side $QR$, line $l$ intersect the sides $PQ$ and $PR$ in two distinct points $M$ and $N$ respectively.

To prove: $\frac{PM}{MQ}=\frac{PN}{NR}$ ... (i)

Construction: $segQN$ and $segRM$ are drawn.

Proof: $\frac{A\left(△PMN\right)}{A\left(△QMN\right)}=\frac{PM}{MQ}$
(Both triangles have equal height with common vertex $M$)

$\therefore \frac{A\left(△PMN\right)}{A\left(△RMN\right)}=\frac{PN}{NR}$ ... (ii)

But $A\left(△QMN\right)=A\left(△RMN\right)$, because they are between parallel lines $MN$ and $QR$ and have equal height corresponding to their common base $MN$ ..... (iii)

From (i), (ii) and (iii), we get

$\frac{A\left(△PMN\right)}{A\left(△QMN\right)}=\frac{A\left(△PMN\right)}{A\left(△RMN\right)}$

$\therefore \frac{PM}{MQ}=\frac{PN}{NR}$ $\left[henceproved\right]$