Solved Example
B
A
M
L
C
BL and CM are medians of a triangle ABC right angled at A. Prove that 4 (BL2 + CM2) = 5 BC2.
Sol:
BC2
=
AB2
+
AC2
(Pythagoras Theorem)
……(2)
In ΔABL,
BL2
=
AL2
+
AB2
BL2
In ΔABC,
=
AC
2
+
AB2
….[From (1)]
(
)
2
CL
= LA
AM
= BM
}
[BL and CM are medians
of a ΔABC]
……(1)
∠A = 900
(Pythagoras Theorem)
∴
∴
∠A = 900
1
2
CA
=
1
2
AB
=
B
A
M
L
C
BL and CM are medians of a triangle ABC right angled at A. Prove that 4 (BL2 + CM2) = 5 BC2.
Sol:
BL2
=
AC2
4
+
AB2
∴
4 BL2
=
AC2
+
4 AB2
∴
……(3)
Solved Example
CM2
=
AB2
4
+
∴
AC2
4 CM2
=
4 AC2
+
AB2
……(4)
In ΔCMA,
CM2
=
AC2
+
AM2
CM2
=
AB
2
+
[From……(1)]
∴
(
)
2
AC2
∴
∠A = 900
CL
= LA
AM
= BM
1
2
CA
=
1
2
AB
=
B
A
M
L
C
BL and CM are medians of a triangle ABC right angled at A. Prove that 4 (BL2 + CM2) = 5 BC2.
Sol:
4BL2
=
AC2
+
4 AB2
∴
……(3)
4CM2
=
4 AC2
+
AB2
……(4)
∴
4 (BL2
=
5 (AC2
+
AB2)
+
CM2)
4 (BL2
=
5
BC2
+
CM2)
[From (1)]
4 BL2
=
AC2
+ 4AB2
+
4CM2
∴
∴
∴
+ 4AB2
+ 4AC2
+ 4AB2
(Adding (3) and (4), we have)
Solved Example
CL
= LA
AM
= BM
1
2
CA
=
1
2
AB
=