Question

If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then prove that two triangles are congruent.

Answer 1

Given:Two right triangles $△ABC$ and $△DEF$ where $\angle B={90}^{\circ }$ and $\angle E={90}^{\circ }$, hypotenuse is equal.

i.e., $AC=DF$ and one side is equal i.e.,$BC=EF$

To prove:$△ABC\cong △DEF$

Proof:In right $△ABC$

By pythagoras theorem, ${AC}^2={AB}^2+{BC}^2$

${AB}^2={AC}^2-{BC}^2$ ........$\left(1\right)$

In right $△DEF$

By pythagoras theorem, ${DF}^2={DE}^2+{EF}^2$

${DE}^2={DF}^2-{EF}^2$ ........$\left(2\right)$

From $\left(1\right)$

${AB}^2={AC}^2-{BC}^2$

$\Rightarrow {AB}^2={DF}^2-{EF}^2$ since $AC=DF$ and $BC=EF$

$\Rightarrow {AB}^2={DE}^2$ from $\left(2\right)$

$\Rightarrow AB=DE$ ......$\left(3\right)$

Given:$BC=EF$,$AC=DF$

$\Rightarrow △ABC\cong △DEF$ by SSS congruence.

Hence proved.