Question

Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of an equilateral triangle described on one of its diagonals.
Or, prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Answer 1

Let ABCD be a square of side a.
Therefore, its diagonal is $\sqrt{2}a$.

Two equilateral triangles Δ ABE and Δ DBF are drawn.
Side of equilateral triangle, ΔABE, described on one of its sides = a
Side of equilateral triangle, ΔDBF, described on one of its diagonals= $\sqrt{2}a$
We know that all the angles of an equilateral triangle measure 60º and all its sides are equal.
Therefore, all equilateral triangles are similar to each other.
Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.
$$\begin{gathered} \frac{\mathrm{Areaof}△\mathrm{ABE}}{\mathrm{Areaof}△\mathrm{DBF}}=\frac{{\displaystyle \frac{\sqrt{3}}{4}}A{B}^2}{{\displaystyle \frac{\sqrt{3}}{4}}B{D}^2} \\ ={\left(\frac{\mathrm{a}}{\sqrt{2}\mathrm{a}}\right)}^2 \\ =\frac{{\mathrm{a}}^2}{2{\mathrm{a}}^2}=\frac{1}{2} \end{gathered}$$

OR
Given: Δ ABC ∼ Δ DEF
To Prove: $\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}=\frac{{\mathrm{AB}}^2}{{\mathrm{DE}}^2}=\frac{{\mathrm{AC}}^2}{{\mathrm{DF}}^2}=\frac{{\mathrm{BC}}^2}{{\mathrm{EF}}^2}$

Construction: $\mathrm{Draw}\mathrm{AL}\perp \mathrm{BC}\mathrm{and}\mathrm{DM}\perp \mathrm{EF}$.

Proof: Since Δ ABC ∼ Δ DEF, it follows that they are equiangular and their sides are proportional.
∴ $\angle$A = $\angle$D, $\angle$B = $\angle$E, $\angle$C = $\angle$F
and $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{DF}}$ ..................(i)
$\mathrm{ar}\left(△\mathrm{ABC}\right)=\left(\frac{1}{2}\times \mathrm{BC}\times \mathrm{AL}\right)$
$\mathrm{ar}\left(△\mathrm{DEF}\right)=\left(\frac{1}{2}\times \mathrm{EF}\times \mathrm{DM}\right)$
$\Rightarrow \frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}=\frac{\left(\frac{1}{2}\times \mathrm{BC}\times \mathrm{AL}\right)}{\left(\frac{1}{2}\times \mathrm{EF}\times \mathrm{DM}\right)}=\frac{\mathrm{BC}}{\mathrm{EF}}\times \frac{\mathrm{AL}}{\mathrm{DM}}$ ...............(ii)
In Δ ALB and Δ DME, we have:
∠ALB = ∠DME = 90° and $\angle$B = $\angle$∠E [ from (i)]
∴$\angle$ALB ∼ $\angle$DME

Consequently, $\frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\mathrm{AB}}{\mathrm{DE}}$
$\mathrm{But}\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}$ [from (i)]

∴ $\frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\mathrm{BC}}{\mathrm{EF}}$..................(iii)
Using (iii) in (ii), we get:
$\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}=\left(\frac{\mathrm{BC}}{\mathrm{EF}}\times \frac{\mathrm{BC}}{\mathrm{EF}}\right)=\frac{{\mathrm{BC}}^2}{{\mathrm{EF}}^2}$
Similarly, $\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}=\frac{{\mathrm{AB}}^2}{{\mathrm{DE}}^2}\mathrm{and}\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}=\frac{{\mathrm{AC}}^2}{{\mathrm{DF}}^2}$

Hence, $\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}=\frac{{\mathrm{AB}}^2}{{\mathrm{DE}}^2}=\frac{{\mathrm{AC}}^2}{{\mathrm{DF}}^2}=\frac{{\mathrm{BC}}^2}{{\mathrm{EF}}^2}$