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

Explanation

`AB= AC and AD = AF rArr AB -AD = AC - AF rArr DB = CF. `
We have `triangle EFC ~= triangle EDB " " [because CF= DB, angle EFC=angle EDB=90^(@), EF=ED].`
`therefore CE=EB " (c.p.c.t.)." `