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

1.	If AL is an aptitude of a triangle ABC, show that AB+AC>2AL ?
Answer» ong>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]

Answer»

ong>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]

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