Question

If the areas of two similar triangles are equal, prove that they are congruent. [3 MARKS]

Answer 1

Concept: 1 Mark
Application: 1 Mark
Proof : 1 Mark

Let$\Delta ABC\backsim \Delta DEF$

Given that,

$Area\Delta ABC=Area\Delta DEF$

Since the ratio of the area of two similair triangles is equal to the ratio of the squares on thier corresponding sides, we have.

$\frac{Areaof\left(\Delta ABC\right)}{Areaof\left(\Delta DEF\right)}=\frac{A{B}^2}{D{E}^2}=\frac{A{C}^2}{D{F}^2}=\frac{B{C}^2}{E{F}^2}$

$\Rightarrow \frac{A{B}^2}{D{E}^2}=\frac{A{C}^2}{D{F}^2}=\frac{B{C}^2}{E{F}^2}=1$ [ $\because Area\Delta ABC=Area\Delta DEF)$]

$\Rightarrow A{B}^2=D{E}^2,A{C}^2=D{F}^2$ and $B{C}^2=E{F}^2$

$\Rightarrow AB=DE,AC=DFandBC=EF$

$\therefore \Delta ABC\cong \Delta DEF$ [By SSS congruence]