A right angled triangle has the vertices (1,1), (1,7) and (7,1) prove that the mid point of the hypotenuse
lies at equal distance from each of its vertices.
Answers
Step-by-step explanation:
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Answer:
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.