Question

if AL is an aptitude of a triangle ABC, show that AB+AC>2AL ?​

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[(AD)²+(BD)²]