Question

If a line parallel to one of the sides of a triangle intersects the other two sides in distinct points, then the segment of the other two sides in one halfplane are proportional to the segments in the other halfplane.

Answer 1

Given: In the plane of $\Delta ABC$, a line $l||BC$, l intersects $\overline{AB}$ and $\overline{AC}$ at points P and Q respectively.
To prove : $\frac{AP}{PB}=\frac{AQ}{QC}$
Proof: Let $\overline{OM}\perp \overline{AB}$ and $\overline{PN}\perp \overline{AC}$. Construct $\overline{BQ}$ and $\overline{CP}$.
Area of a triangle $=\frac{1}{2}\times base\times altitude$
$\therefore$ Area of $\Delta APQ=\frac{1}{2}\times AP\times QM$
Area of $\Delta PBQ=\frac{1}{2}\times PB\times QM$
$\therefore \frac{Areaof\Delta APQ}{Areaof\Delta PBQ}=\frac{\frac{1}{2}\times AP\times QM}{\frac{1}{2}\times PB\times QM}=\frac{AP}{PB}$ ..... (i)
Also area of $\Delta APQ=\frac{1}{2}\times AQ\times PN$
Area of $\Delta CPQ=\frac{1}{2}\times QC\times PN$
$\therefore \frac{Areaof\Delta APQ}{Areaof\Delta CPQ}=\frac{\frac{1}{2}\times AQ\times PN}{\frac{1}{2}\times QC\times PN}=\frac{AQ}{QC}$ .... (ii)
$\Delta PBQ$ and $\Delta PCQ$ are having common base $\overline{PQ}$ and they are lying between two parallel lines $\overleftrightarrow{PQ}$ and $\overleftrightarrow{BC}$.
Area of $\Delta PBQ$ = Area of $\Delta PCQ$ ......... (iii)
From (i), (ii) and (iii) $\frac{AP}{PB}=\frac{AQ}{QC}$