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Question

(1)
In Figure, AD is a median of a triangle ABC and AMBC.
Prove that :
(i)AC2=AD2+BC×DM+(BC2)2



(2)
In Figure, AD is a median of a triangle ABC and AMBC.
Prove that : (ii)AB2=AD2BC×DM+(BC2)2
(3)
In Figure, AD is a median of a triangle ABC and AMBC.
Prove that : (iii)AC2+AB2=2AD2+12BC2

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Solution


(1)

Using Pythagoras Therom in ΔAMD & ΔAMC
In ΔAMD
On applying Pythagoras theorem,
AD2=AM2+MD2.......(1)

In ΔAMC
On applying Pythagoras theorem,
AC2=AM2+MC2
AC2=AM2+(MD+DC)2
AC2=AM2+MD2+DC2+2(MD)(DC)

Replacing MC in ΔAMC
In ΔAMC
AC2=AM2+MC2
AC2=AM2+(MD+DC)2
AC2=AM2+MD2+DC2+2(MD)(DC)
From (1)
AC2=AD2+DC2+2(MD)(DC)......(2)

As AD is median, D is the midpoint of BC
Substitude DC=BC2 in (2)
AC2=AD2+(BC2)2+(MD)(BC)
Hence, proved.

(2)

Using Pythagoras Theorem in ΔAMB and ΔAMD
In ΔAMB
On applying Pythagoras theorem,
AB2=AM2+BM2.....(1)

In ΔAMD
On applying Pythagoras theorem,
AD2=AM2+MD2......(2)

From (1) and (2), equation (1) can be written as,
AB2=(AD2DM2)+BM2
AB2=(AD2DM2)+(BDDM)2 (\because BM=BD-DM)
AB2=AD2DM2+BD2+DM22(BD)(DM)
AB2=AD2+BD22(BD)(DM).....(1)

On substituting BD
Substitute BD=BC2 in (3)
AB2=AD2BC×DM+(BC2)2
Hence Proved.

(3)

Applying Pythagoras Theorem in ΔAMB and ΔAMC
In ΔAMB
On applying Pythagoras theorem,
AB2=AM2+BM2.....(1)

In ΔAMC
On applying Pythagoras theorem,
AC2=AM2+MC2........(2)

Adding equation (1) &(2).
AC2+AB2=2AM2+BM2+MC2
Substitute BM=BD-DM and MC=MD+DC
AC2+AB2=2AM2+(BDDM)2+(MD+DC)2
AC2+AB2=2AM2+BD2+DM22(BD)(DM)+MD2+DC2+2(MD)(DC)

Substituting BD and DC SubstitudeBD=DC=BC2 in the above equation,

AC2+AB2=2AM2+(BC2)2+2DM22(BC2)(DM)+MD2+(BC2)2+2(MD)(BC2)

AC2+AB2=AD2+(BC2)2+AM2(BC)(DM)+(MD+DC)2

AC2+AB2=AD2+(BC2)2+AM2(BC)(DM)+(CM)2

AC2+AB2=AD2+(BC2)2+AC2(BC)(DM)
AB2=AD2+(BC2)2(BC)(DM)

Hence Proved.

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