Question

In given figure prove that the area of the triangle BCE described on one side BC of a square ABCD as base is one half the area of the similar triangle ACF described on the diagonal AC as base.

Answer 1

ABCD is a square. $△BCE$ is described on side BC is similar to $△ACF$ desctibed on diagonal AC.
Since ABCD is a square. Therefore,

$AB=BC=CD=DA$ and $AC=\sqrt{2}BC$ [$\because Diagonal=\sqrt{2}\left(side\right)$]

Now, $△BCE\sim △ACF$

$\Rightarrow$ $\frac{Area\left(△BCE\right)}{Area\left(△ACF\right)}=\frac{{BC}^2}{{AC}^2}$

$\Rightarrow$ $\frac{Area\left(△BCE\right)}{Area\left(△ACF\right)}=\frac{{BC}^2}{{\left(\sqrt{2}BC\right)}^2}=\frac{1}{2}$

$\Rightarrow$ $Area\left(△BCE\right)=\frac{1}{2}Area\left(△ACF\right)$ $\left[Henceproved\right]$