AD is the altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed. Prove that area ADE: area ABC=3:4

since both triangles are equilateral, all of their angles are equal to 60 degree.

this implies that both triangles are similar.( by AAA similarity criterion)

=> ABC ~ ADE

since ABC is an equilateral triangle, the altitude AD divides BC into two equal parts, forming the right triangles ADC and ADB .

now, let the length of each side of ABC be equal to @,

then, BD = DC = @/2

in triangle ADC, by pythagoras theorem,

AC² = AD² + DC²

=> @² = AD² + @²/4

=> AD² = 3@²/4

=> AD = root 3 @/ 2

=> @/AD = 2/root 3

=> AB/AD = 2/root 3

also, from (1),

ABC ~ ADE

we know that the ratio of areas of two similar triangles is equal to the square of any two proportional sides of the triangles ,

therefore, ar. ADE: ar. ABC = AD²:AC² = 2²:(root 3)²

= 4: 3

or, ar.ABC : ar.ADE = 3:4

hence,proved.