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.

Answer 1

Given: In isosceles triangle ABC, a square ΔDEF is inscribed.

To prove: CE = BE

Proof: In an isosceles ΔABC,

∠A = 90°

And AB = AC  ...(i)

Since, ΔDEF is a square.

AD = AF  [All sides of square are equal] ...(ii)

On subtracting equation (ii) from equation (i), we get

AB – AD = AC – AF

BD = CF  ...(iii)

Now, in ΔCFE and ΔBDE,

BD = CF  ...[From equation (iii)]

DE = EF   ...[Sides of a square]

And ∠CFE = ∠EDB   ...[Each 90°]

ΔCFE ≅ ΔBDE   ...[By SAS congruence rule]

∴ CE = BE   ...[By CPCT]

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