Question

If the area of two similar triangles is equal, prove that they are congruent. $$$$

Answer 1

Hint: We take two similar triangles $\mathrm{\Delta }ABC\sim \mathrm{\Delta }DEF$ and where ratios of sides are ${\displaystyle \frac{AB}{DE}}={\displaystyle \frac{BC}{EF}}={\displaystyle \frac{AC}{DF}}$ . We use the theorem that the ratio of areas of two similar triangles is equal to the squared ratio of corresponding sides to prove $AB=DE,BC=EF,AC=DF$ which means the triangles are congruent. $$$$

Complete step by step answer:

Let us take two similar triangles $\mathrm{\Delta }ABC\sim \mathrm{\Delta }DEF$ . We know that in similar triangles the measures of corresponding angles are equal and the sides opposite to them are called corresponding sides. The ratios of corresponding sides are equal. So let us take $\mathrm{\angle }A=\mathrm{\angle }D,\mathrm{\angle }B=\mathrm{\angle }E,\mathrm{\angle }C=\mathrm{\angle }F$ and have the ratios of corresponding sides
$${\displaystyle \frac{AB}{DE}}={\displaystyle \frac{BC}{EF}}={\displaystyle \frac{AC}{DF}}$$
We know the theorem that if two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides. So we have
$${\displaystyle \frac{\text{Area of }\mathrm{\Delta }\text{ABC}}{\text{Area of }\mathrm{\Delta }DEF}}={\left({\displaystyle \frac{AB}{DE}}\right)}^2={\left({\displaystyle \frac{BC}{EF}}\right)}^2={\left({\displaystyle \frac{AC}{DF}}\right)}^2$$
We are given in the question that $\text{Area of }\mathrm{\Delta }\text{ABC}=\text{Area of }\mathrm{\Delta }DEF$ . So we have;
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{\text{Area of }\mathrm{\Delta }\text{ABC}}{\text{Area of }\mathrm{\Delta }\text{ABC}}}={\left({\displaystyle \frac{AB}{DE}}\right)}^2={\left({\displaystyle \frac{BC}{EF}}\right)}^2={\left({\displaystyle \frac{AC}{DF}}\right)}^2\\ & \Rightarrow 1={\left({\displaystyle \frac{AB}{DE}}\right)}^2={\left({\displaystyle \frac{BC}{EF}}\right)}^2={\left({\displaystyle \frac{AC}{DF}}\right)}^2\end{array}$$
Let us consider
$$1={\left({\displaystyle \frac{AB}{DE}}\right)}^2$$
We take square root both sides to have
$$\begin{array}{rl}& \Rightarrow 1={\displaystyle \frac{AB}{DE}}\\ & \Rightarrow AB=DE.......\left(1\right)\end{array}$$
Similarly we consider
$$1={\left({\displaystyle \frac{BC}{EF}}\right)}^2$$
We take square root both sides to have
$$\begin{array}{rl}& \Rightarrow 1={\displaystyle \frac{BC}{EF}}\\ & \Rightarrow BC=EF.......\left(2\right)\end{array}$$
 Similarly we consider
$$1={\left({\displaystyle \frac{AC}{DF}}\right)}^2$$
We take square root both sides to have
$$\begin{array}{rl}& \Rightarrow 1={\displaystyle \frac{AC}{DF}}\\ & \Rightarrow AC=DF.......\left(3\right)\end{array}$$
We conclude from equations (1), (2) and (3) that
$$AB=DE,BC=EF,AC=DF$$
Since the lengths of the corresponding sides of the triangles are equal by side-side-side criteria the triangles are congruent which means $\mathrm{\Delta }ABC\cong \mathrm{\Delta }DEF$ . Hence the statement is proved. $$$$



Note:
We note that all congruent triangles are similar but not all similar triangles are congruent and both for similar and congruent triangles the measures of corresponding triangles are equal. We also note the length of sides is always positive and hence we have rejected the negative square root here.