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Proof: In right angled ΔABC
By Pythagoras theorem
AC
2
=AB
2
+BC
2
.......... (1)
In right angled ΔBMC
By Pythagoras theorem
CM
2
=BM
2
+BC
2
CM
2
=(
2
1
AB)
2
+BC
2
[M is the mid-point of AB]
CM
2
=
4
1
AB
2
+BC
2
4CM
2
=AB
2
+4BC
2
......... (2)
(Multiplying by 4 on both sides)
In right angled ΔABN
By Pythagoras theorem
AN
2
=AB
2
+BN
2
AN
2
=AB
2
+(
2
1
BC)
2
AN
2
=AB
2
+
4
1
BC
2
∴4AN
2
=4AB
2
+BC
2
........... (3)
(Multiplying by 4 on both sides)
Adding (2) and (3)
∴4CM
2
+4AN
2
=AB
2
+4BC
2
+4AB
2
+BC
2
∴4(CM
2
+AN
2
)=5AB
2
+5BC
2
∴4(CM
2
+AN
2
)=5(AB
2
+BC
2
)
∴4(CM
2
+AN
2
)=5AC
2
∴4(AN
2
+CM
2
)=5AC
2
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