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$AD$ is an altitude of an equilateral triangle $ABC$. On $AD$ as base, another equilateral triangle $ADE$ is constructed. Prove that Area($\triangle ADE$):Area($\triangle ABC$)$=3:4$.
Given:
$AD$ is an altitude of an equilateral triangle $ABC$. On $AD$ as base, another equilateral triangle $ADE$ is constructed.
To do:
We have to prove that Area($\triangle ADE$):Area($\triangle ABC$)$=3:4$.
Solution:
Let $AB=BC=AC=2x$
In an equilateral triangle altitude is also the perpendicular bisector.
This implies,
$BD=DC=x$
$\triangle ADB$ is a right triangle. Therefore, by using Pythagoras theorem,
$AB^2=AD^2+BD^2$
$(2x)^2=AD^2+x^2$
$AD^2=4x^2-x^2$
$AD^2=3x^2$
$AD=\sqrt{3x^2}$
$AD=\sqrt3x$
$\triangle ABC$ and $\triangle ADE$ are equilateral triangles and hence equiangular.
Therefore,
$\triangle ABC \sim\ \triangle ADE$
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.
This implies,
$\frac{ar(\triangle ADE)}{ar(\triangle ABC)}=\frac{AD^2}{BC^2}$
$=\frac{3x^2}{(2x)^2}$
$=\frac{3x^2}{4x^2}$
$=\frac{3}{4}$
Hence proved.
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