Consider the triangle ABC.
If we draw a line PQ parallel to BC according to the triangle proportionality theorem the ratio of AP to PB is equal to the ratio of AQ to QC
We are using the two-column proof method.
Join the vertex B and Q and also Join the vertex P and C.
Draw a perpendicular line from Q to AB , and mark the point as N.
Draw a perpendicular line from P to AC , and mark the point as M.
[tex]S.no\ldots..\text{Statement }\ldots\ldots\ldots\ldots..\ldots...\operatorname{Re}ason[/tex][tex]1\ldots..Area\text{ of }\Delta APQ=\frac{1}{2}\times AP\times QN\ldots\ldots\ldots\ldots\text{.}area\text{ of triangle=}\frac{1}{2}\times base\times height.[/tex]QN is the height of the triangle PBQ and APQ.
[tex]2\ldots..Area\text{ of }\Delta PBQ=\frac{1}{2}\times AB\times QN\ldots\ldots\ldots\ldots\text{.}area\text{ of triangle=}\frac{1}{2}\times base\times height.[/tex]PM also the height of the triangle APQ.
[tex]3\ldots..Area\text{ of }\Delta APQ=\frac{1}{2}\times AQ\times PM\ldots\ldots\ldots\ldots\text{.}area\text{ of triangle=}\frac{1}{2}\times base\times height.[/tex][tex]4\ldots..Area\text{ of }\Delta QCP=\frac{1}{2}\times QC\times PM\ldots\ldots\ldots\ldots\text{.}area\text{ of triangle=}\frac{1}{2}\times base\times height.[/tex][tex]5\ldots..\frac{Area\text{ of }\Delta APQ}{Area\text{ of }\Delta PBQ}=\frac{\frac{1}{2}\times AP\times QN}{\frac{1}{2}\times AB\times QN}=\frac{AP}{AB}\ldots\ldots\ldots\ldots\text{.from (1) and (2)}[/tex][tex]6\ldots..\frac{Area\text{ of }\Delta APQ}{Area\text{ of }\Delta\text{QCP}}=\frac{\frac{1}{2}\times AQ\times PM}{\frac{1}{2}\times QC\times PM}=\frac{AQ}{QC}\ldots\ldots\ldots\ldots\text{.from (3) and (4)}[/tex][tex]7\ldots..Area\text{ of }\Delta PBQ=Area\text{ of }\Delta QCP\ldots\ldots\ldots\ldots\text{.}By\text{ known theorem, which stat}es\text{ that triangle drawn betwe}en\text{ two parallel lines and on the same base have equal areas}.[/tex][tex]8\ldots..\frac{AP}{PB}=\frac{AQ}{QC}\ldots\ldots\ldots\ldots\text{.from (5), (6) and (7)}[/tex]Hence proved the theorem.
