Question

Prove that the mid-point of the hypotenuse of a right-angled triangle is equidistant from all the vertices.

Answer 1

Let ABC be a right triangle, righte angled at A. Let D be the midpoint of the hypotenuse BC. We have to show that AD = CD = BD. Now it is obvious that CD = BD = 1212 BC. Since D is the midpoint of BC.

consider,

AD

=

AB

+

BD

=

AB

+

2

1

BC

=

AB

+

2

1

(

BA

+

AC

)

AD

=

AB

−

2

1

AB

+

2

1

AC

=

2

1

(

AB

+

AC

)

∴(

AD

)

2

=

4

1

(

AB

+

AC

)

2

(

AD

)

2

=

4

1

(

AB

+

AC

)

2

=

4

1

(

AB

2

+

AC

2

+2

AB

.

AC

)

i.e. , AD

2

=

4

1

(AB

2

+AC

2

+0)since(

AB

⊥

AC

)

=AD

2

=

4

1

BC

2

AD=

2

1

BC

So we have AD = BD = CD

Hence proved.