AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed.
Prove that Area (∆ADE) : Area (∆ABC) = 3 : 4.
We have an equilateral triangle $\triangle A B C$ in which $\mathrm{AD}$ is altitude. An equilateral triangle $\triangle A D E$ is drawn using $\mathrm{AD}$ as base. We have to prove that, $\frac{a r(\Delta A D E)}{a r(\Delta A B C)}=\frac{3}{4}$
Since the two triangles are equilateral, the two triangles will be similar also.
$\triangle A D E-\triangle A B C$
We know that according to the theorem, the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.
$\frac{\operatorname{ar}(\Delta A D E)}{\operatorname{ar}(\Delta A B C)}=\left(\frac{A D}{A B}\right)^{2}$....(1)
Now $\triangle A B C$ is an equilateral triangle. So,
$\angle B=60^{\circ}$
Therefore,
$\sin \angle B=\frac{A D}{A B}$
So, $\frac{A D}{A B}=\frac{\sqrt{3}}{2}$
We will now use this in equation (1). So,
$\frac{\operatorname{ar}(\Delta A D E)}{\operatorname{ar}(\Delta A B C)}=\left(\frac{A D}{A B}\right)^{2}$
$=\left(\frac{\sqrt{3}}{2}\right)^{2}$
$=\frac{3}{4}$
Hence, proved.