If the mean of abc is equal
to median of a, b, c then a<b <c
then show that mean of ac
is equal to median of a, b, c.
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AD is a median of △ABC.
⇒ Draw AE⊥BC.
In right angled △AEB,
⇒ (AB)
2
=(AE)
2
+(BE)
2
[ By Pythagoeas theorem ] --- ( 1 )
In right angled △ACE,
⇒ (AC)
2
=(AE)
2
+(EC)
2
[ By Pythagoeas theorem ] ---- ( 2 )
Adding ( 1 ) and ( 2 ),
⇒ (AB)
2
+(AC)
2
=(AE)
2
+(BE)
2
+(AE)
2
+(EC)
2
⇒ (AB)
2
+(AC)
2
=2(AE)
2
+(BD−ED)
2
+(ED+DC)
2
⇒ (AB)
2
+(AC)
2
=2(AE)
2
+(BD)
2
−2BD.ED+(ED)
2
+(ED)
2
+2ED.DC+(DC)
2
⇒ (AB)
2
+(AC)
2
=2(AE)
2
+2(ED)
2
+(BD)
2
+(DC)
2
[ Since, BD=DC ]
⇒ (AB)
2
+(AC)
2
=2(AE)
2
+2(ED)
2
+2(BD)
2
[ Since, BD=DC ]
⇒ (AB)
2
+(AC)
2
=2[(AE)
2
+(ED)
2
+(BD)
2
]
⇒ (AB)
2
+(AC)
2
=2[(AD)
2
+(BD)
2
]
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