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.

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.

Philoid · Accepted Answer

To Prove: The vertex of the square opposite the vertex of the common angle bisects the hypotenuse.

Given: A square is inscribed in an isosceles right-angled triangle with one angle common.

Concept Used:

SAS Congruency: If two sides and one angle of a triangle is equal to two sides and one angle of another triangle then both the triangles are called congruent.

In square all sides are equal.

Diagram:

Proof:

Let the square be DEFB

Therefore, DE = EF = FB = BD

ΔABC is isosceles so AB = BC

Therefore,

And as DB = FC, we can say that,

AD = FC

Now consider ΔADE and ΔEFC

AD = FC

DE = EF [Sides of square]

∠ADE = ∠EFC

So, both triangles are congruent.

AE = EC by C.P.C.T.

Hence, Proved.