Question

Prove that "If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side".

Answer 1

Given : The line $l$ intersects the sides $PQ$ and side $PR$ of $\Delta PQR$ in the points $M$ and $N$ respectively such that $\frac{PM}{MQ}=\frac{PN}{NR}$ and $P-M-Q$, $P-N-R$.
To Prove : Line $l$ $\parallel$ Side $QR$
Proof : Let us consider that line $l$ is not parallel to the side $QR$. Then there must be another line passing through $M$ which is parallel to the side $QR$.
Let line $MK$ be that line.
Line $MK$ intersects the side $PR$ at $K$, $(P-K-R)$
In $\Delta PQR$, line $MK\parallel$ side $QR$
$\therefore$ $\frac{PM}{MQ}=\frac{PK}{KR}$ ....(1) (B.P.T.)
But $\frac{PM}{MQ}=\frac{PN}{NR}$ ....(2) (Given)
$\therefore$ $\frac{PK}{KR}=\frac{PN}{NR}$ [From (1) and (2)]
$\therefore$ $\frac{PK+KR}{KR}=\frac{PN+NR}{NR}$ ($P-K-R$ and $P-N-R$)
$\therefore$ the points $K$ and $N$ are not different.
$\therefore$ line $MK$ and line $MN$ coincide
$\therefore$ line $MN\parallel$ Side $QR$
Hence, the converse of B.P.T. is proved.