Question

Prove that in a triangle the sum of any two medians is greater than the third median.

Answer 1

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 > BE.

Answer 2

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

We have to prove that: (1) AD + BE > CF

(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

ΔABH, F is the mid-point of AB and G is the ∴FG||BH (Mid-point theorem)

So, FG || BH (Mid-point theorem)

Hense GC || BH ...............1

Similarly, BG || HC ...........2

From 1 and 2, we get

BGCH is a parallelogram (Both pair of opposite sides) So, BH = CG (3) (Opposite sides of parallelogram are In ΔBGH,

So, BH = CG ............3 (Opposite sides of parallelogram are equal)

In ΔBGH,

BG + GH > BH (Sum of any two 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 > BE

HENCE PROVED

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