Question

If in two triangle,hypotenuse and one side of a triangle are equal to the hypotenuse and one side of other triangle, prove that the two triangle are congruent ?

Answer 1

Given, In $△ABC$ and $△DEF$ the hypotenuse and one side of the triangle are equal to the hypotenuse and one side of the other triangle. So, $△ABC$ and $△DEF$ are right-angled triangles.

Let, $AC$ and $DF$ are the two equal hypotenuses and $BC$ and $EF$ are the two equal sides of $△ABC$ and $△DEF$ respectively i.e. $AC=DF,BC=EF$.

In $△ABC$,

$A{C}^2={AB}^2+{BC}^2\Rightarrow {AB}^2=A{C}^2-{BC}^2⟶\left(1\right)$

In $△DEF$,

${DF}^2={DE}^2+E{F}^2\Rightarrow {DE}^2={DF}^2-E{F}^2⟶\left(2\right)$

From equation $\left(1\right)$,

${AB}^2=A{C}^2-{BC}^2\Rightarrow {AB}^2={DF}^2-E{F}^2[\because AC=DF,BC=EF]\therefore AB=DE⟶\left(3\right)$

$\therefore$ in $△ABC$,

$AB=DE\left[from\right(3\left)\right]AC=DF\left[Given\right]BC=EF\left[Given\right]\Rightarrow △ABC\cong △DEF\left[bySSScongruencerule\right]$

Hence, proved.