Question

Prove that if the areas of two similar triangle are equal , then the triangles are congruent.

Answer 1

Area of a triangle given its $2$ sides and an angle included is $=\frac{1}{2}\times AB\times AC\times \sin \theta$

For $△DEF,$ area $=\frac{1}{2}\times DE\times DF\times \sin \varphi$

given area$\left(△ABC\right)=$ar$\left(△DEF\right)$

$\Rightarrow \frac{1}{2}\times AB\times AC\times \sin \theta =\frac{1}{2}\times DE\times DF\times \sin \varphi$

This means that corresponding sides are equal and included angles are also equal.

i.e.,$AB=DE,AC=DF$

and $\theta =\varphi \Rightarrow \angle BAC=\angle EDF$

$\therefore$ By $SAS$ congruence the two triangles are congruent .