Question

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.​

Answer 1

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

]

Answer: