In a triangle ABC, the sum of two medians is greater than third one:

1.	In a triangle ABC, the sum of two medians is greater than third one:
Answer» 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. thanks

Answer»

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.

thanks

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