If the areas of two similar triangles are equal, prove that they are congruent.

Given:

The areas of two similar triangles are equal.

To do:

We have to prove that they are congruent.

Solution:

Let $\triangle \mathrm{ABC}  \sim \Delta \mathrm{DEF}$ such that  $ar(\triangle \mathrm{ABC})=ar(\Delta \mathrm{DEF})$

$\frac{ar(\triangle \mathrm{ABC})}{ar(\triangle \mathrm{DEF)}}=\frac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}$

$=\frac{\mathrm{AC}^{2}}{\mathrm{DF}^{2}}$

$=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}$

This implies,

$\frac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}=\frac{\mathrm{AC}^{2}}{\mathrm{DF}^{2}}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}=1$

Therefore,

$\mathrm{AB}^2=\mathrm{DE}^{2}$

$\mathrm{AB}=\mathrm{DE}$

$\mathrm{AC}^2=\mathrm{DF}^{2}$

$\mathrm{AC}=\mathrm{DF}$

$\mathrm{BC}^{2}=\mathrm{EF}^{2}$

$\mathrm{BC}=\mathrm{EF}$

Therefore, by SSS criterion,

$\triangle \mathrm{ABC} \cong \Delta \mathrm{DEF}$

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Updated on: 10-Oct-2022

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