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.

Explanation

Let we have an equilateral `/_\ABC` in which AD is altitude. An equilateral `/_\ADE` is drawn using AD as base.

Since, the two triangle are equilateral, the two triangles will be similar also.

`/_\ABC~= /_\ADE`

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.

(Area of` /_\ADE`)/(Area of` /_\ABC`)=` ((AD)/(AD))^2 ` ...eq(1)

In equilateral `/_\ABC`

`/_B = 60^0`

` sin B = (AD)/(AB) `

...