Prove that the midpoint of the hypotenuse of a right angled triangle is equidistant from the vertices
Answers
Explanation:
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
Let P be the mid point of the hypo. of the right triangle ABC, right angled at B.
Draw a line parallel to BC from P meeting AB at D.
Join PB.
in triangles,PAD and PBD,
angle PDA= angle PDB (90 each due to conv of mid point theorem)
PD=PD(common)
AD=DB( as D is mid point of AB)
so triangles PAD and PBD are congruent by SAS rule.
PA=PB(C.P.C.T.)
but
PA=PC(given as P is mid point )
So,
PA=PC=PB
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