A square is incribed in an isoceles right triangle, so that the square and the triangle have one angle common. Show that the vertex of the sqare opposite the vertex of the common angle bisects the hypotenuse.

Explanation

Given In isosceles traingle ABC, a square ADEF is inscribed.
To prove `" "` CE=BE
Proof In an isosceles `Delta`ABC, `angleA=90^(@)`
and `" "AB=AC" "…(i)`
Since, ADEF is a square.
`therefore" "AD=AF" "`[all sides of square are equal]...(ii)
On subtracting Eq. (ii) from Eq. (i), we get
`" "AB-AD=AC-AF`
`rArr" "BD=CF" "...(iii)`
Now, in `DeltaCFE and DeltaBDE,`

`" "BD = CF" "`[from Eq. (iii)]
`" "DE=EF" "`[sides of a square]
and `" "anglefCFE=angleEDB" "` [each `90^(@)`]
`" "Delta CFE~=DeltaBDE" "` [by SAS congruence rule]
`therefore" " CE=BE " "` [by CPCT]
Hence, vertex E of the square bisects the hypotenuse BC.