Question

Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
Or, prove that in a triangle, if the square of one side is equal to the sum of the squares of the other two sides, the angle opposite the first side is a right angle.

Answer 1

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

OR

Given: In Δ ABC, AC² = AB² + BC²
To prove: $\angle$B = 90°
Construction: Draw Δ DEF, such that DE = AB, EF = BC and $\angle$E = 90°.

Proof:
In Δ DEF, we have $\angle$E = 90°
So, by Pythagoras' theorem, we have:
DF² = DE² + EF²
⇒ DF² = AB² + BC² ..............(i) [Since DE = AB and EE = BC]
But, AC² = AB² + BC² .............(ii) [given]

From (i) and (ii), we get:
AC² = DF² ⇒ AC = DF
Now in Δ ABC and Δ DEF, we have:
AB = DE, BC = EF and AC = DF
∴ Δ ABC ∼ Δ DEF
Hence, $\angle$B=$\angle$E = 90°.