Question

prove that in a triangle ABC sum of two medians is greater than third medians​

Answer 1

Step-by-step explanation:

Given: AD, BE and CF are the medians of ΔABC.

 

To prove: (1) AD + BE > CF

  (2) BE + CF > AD

  (3) AD + CF > BE

 

Construction: Produce AD to H, such that AG = GH.

Join BH and CH



Proof: In ΔABH, F is the mid-point of AB and G is the mid point of AH.

∴FG||BH  (Mid-point theorem)

∴ GC||BH    (1)

 

Similarly, BG||HC  (2)

From (1) and (2), we get

BGCH is a parallelogram  (Both pair of opposite sides are parallel)

∴ BH = CG  (3)  (Opposite sides of parallelogram are equal)

 

In ∆BGH,

BG + GH > BH  (Sum of any to sides of a triangle is always greater than the third side)

⇒ BG + AG > CG  (GH = AG and BH = CG)



Similarly, BE + CF > AD and AD + CF > B