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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.

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Solution.
Given : Here ABC is an isosceles triangle and ADEF is a square inscribed in \triangle ABC.
To prove : CE = BE

Proof: In isosceles \triangle ABC and AB = AC    …..(1)
\angle A=90^{\circ}
Here ADEF is a square

 AD = AF                       …..(2)
Subtract equation 2 from 1
AB - AD = AC - AF
BD = CF                            …..(3)
Now in
\triangle BDE and \triangle CFE
BD = CF           {from equation 3}
\angle CFE = \angle EDB            {90^{\circ} each}
DE = EF                          {side of a square}
\triangle CFE\cong \triangle BDE           {SAS congruence rule}

CE = BE                    {by CPCT}

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

 

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