Question

Prove that the midpoint of the hypotenuse of a right angled triangle is equidistant from all its three 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 =
1
2

BC. Since D is the midpoint of BC.
Consider
→
AD
=
→
AB
+
→
BD
=
→
AB
+
1
2

→
BC
=
→
AB
+
1
2

(
→
BA
+
→
AC
)
=
→
AB
−
1
2

→
AB
+
1
2

→
AC
=
1
2

(
→
AB
+
→
AC
)
∴(
→
AD
)2=
1
4

(
→
AB
+
→
AC
)2=
1
4

(
→
AB
2+2
→
AB
.
→
AC
+
→
AC
2)
i.e., AD2=
1
4

[AB2+0+AC2][sinceAB⊥AC]
=
1
4

BC2 ( by pythagoras theorem)
∴AD=
1
2

BC
So we have AD = BD = CD

Answer 2

Prove that the midpoint of the hypotenuse of a right angled triangle is equidistant from all its three vertices.