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$ABC$ is a triangle and $PQ$ is a straight line meeting $AB$ in $P$ and $AC$ in $Q$. If $AP = 1\ cm, PB = 3\ cm, AQ = 1.5\ cm, QC = 4.5\ cm$, prove that area of $\triangle APQ$ is one-sixteenth of the area of $\triangle ABC$.
Given:
$ABC$ is a triangle and $PQ$ is a straight line meeting $AB$ in $P$ and $AC$ in $Q$.
$AP = 1\ cm, PB = 3\ cm, AQ = 1.5\ cm, QC = 4.5\ cm$.
To do:
We have to prove that area of $\triangle APQ$ is one-sixteenth of the area of $\triangle ABC$.
Solution:
$\frac{PA}{AQ}=\frac{1}{1.5}=\frac{2}{3}$
$\frac{BA}{AC}=\frac{3+1}{1.5+4.5}=\frac{4}{6}=\frac{2}{3}$
In $\triangle APQ$ and $\triangle ABC$,
$\angle PAQ=\angle BAC$ (Common angle)
$\frac{PA}{AQ}=\frac{BA}{AC}$
Therefore,
$\triangle APQ \sim\ \triangle ABC$ (By SAS similarity)
We know that,
If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.
Therefore,
$\frac{ar(\triangle APQ)}{ar(\triangle ABC)}=\frac{(AP)^2}{(AB)^2}$
$=\frac{1^2}{(1+3)^2}$
$=\frac{1}{16}$
This implies,
$ar(\triangle APQ)=\frac{1}{16}ar(\triangle ABC)$
Hence proved.
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