Question

Theorem: If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.

Answer 1

Given : In $△ABC$ line $l\parallel$ line $BC$ and line $l$ intersects $AB$ and $AC$ in point $P$ and $Q$ respectively.
To prove : $\frac{AP}{PB}=\frac{AQ}{QC}$
Construction: Draw seg $PC$ and seg BQ
Proof: $△APQ$ and $△PQB$ have equal heights.
$\frac{A\left(△APQ\right)}{A\left(△PQB\right)}=\frac{AP}{PB}\dots \dots ..$ (I) (areas proportionate to bases)
and $\frac{A\left(△APQ\right)}{A\left(△PQC\right)}=\frac{AQ}{QC}\dots \dots ...$ (II) (areas proportionate to base)
seg $PQ$ is common base of $△PQB$ and $△PQC$.
seg PQ || seg BC, hence D PQB and D PQC have equal heights. $A\left(\Delta PQB\right)=A\left(\Delta PQC\right)\dots \dots ...$ (III)
$\frac{A\left(△APQ\right)}{A\left(△PQB\right)}=\frac{A\left(△APQ\right)}{A\left(△PQC\right)}\dots \dots \dots [$ from (I),$\left(II\right)$ and (III)
$\frac{AP}{PB}=\frac{AQ}{QC}\dots \dots ...[$ from $\left(I\right)$ and $\left(II\right)]$