Y=3(50)x
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Answer:
Given: Consider a triangle ABC in which AD is median drawn from vertex A.
To prove: AB + AC > AD
Proof: In Δ A B D
AB + B D> AD ⇒[In a triangle sum of lengths of two sides is greater than the third side] .................(1)
In Δ A CD
AC + DC > AD [In a triangle sum of lengths of two sides is greater than the third side] .................(2)
Adding (1) and (2)
AB + AC+ B D + D C > B D + DC
A B+ A C+ B C > 2 B D .................(3)
Also , Considering Δ AB C
AB + B C > B C ⇒[In a triangle sum of lengths of two sides is greater than the third side]
⇒ AB + B C - B C >0 ........................(4)
Adding (3) and (4)
A B+ B C+B C+ AB +A C- B C > 2 A D
⇒2 AB + 2 A C> 2 A D
Dividing both side of inequality by 2, we get
A B+ A C> A D
Hence proved.
