Question

A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.

$△ABC$

Answer 1

Given

$∆ABC$ with $\angle A=90°$ and

$∆ABC$ is an isosceles triangle,

To Prove

The vertex of the square opposite the vertex of the common angle bisects the hypotenuse.

vertex $E$ of the square bisect the hypotenuse $BC$

Proof

$∆ABC$ with $\angle A=90°$ and

Since $,ABC$ is an isosceles triangle,

We obtain

$AB=AC-----\left(i\right)$

Let $ADEF$ be the square inscribed in the isosceles triangle $ABC.$.

Then, we have,

$AD=AF=EF=AD----\left(i\right)$

Subtracting equation $\left(ii\right)$ from $\left(i\right),$

$AB-AD=AC-AF$

$BD=CF$

Now,

From $△CFEand△EDB,$

$BD=CF$

$DE=EF$

$\angle CFE=\angle EDB=90°$ (Since, they are the side of a square)

$△CEF~∆BED$

Hence, $CE=BE$

Therefore, vertex $E$ of the square bisect the hypotenuse $BC.$