Question

Question 1
A square is inscribed in an isosceles right triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.

Answer 1

Given in isosceles triangle ABC, a square ADEF is inscribed.

To prove CE = BE

Proof in an isosceles $\Delta ABC,\angle A={90}^{\circ }$

And AB = AC …(i)

Since, ADEF is a square.

$\therefore AD=EF$ [All sides of square are equal …(ii)

On Subtracting Eq. (ii) from Eq. (i) we get

AB - AD = AC – AF

$\Rightarrow BD=CF$ …(iii)

Now, in $\Delta CFE\text{and}\Delta BDE$,

BD = CF [From eq. (iii)]

DE = EF [ Sides of a square]

And $\angle CFE=\angle EDB$ [ each ${90}^{\circ }$]

$\Delta CFE\cong \Delta BDE$ [ by SAS congruence rule]

$\therefore CE=BE$ [ by CPCT]

Hence, vertex E of the square bisects the hypotenuse BC.