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.

Name: 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.
Uploaded: Thu, 2021-01-07 22:11
Description: 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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A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle in common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.

Answers (1)

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