Question

in ∆ABC ,if AD is the median drawn to BC then AB²+AC²/AD²+CD²=​

Answer 1

Answer:

⇒  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]