Question

the sum of any 2 sides of a triangle is greater than the median of the third

Answer 1

${\blue{\mathfrak{\underline{\underline{Solution:-}}}}}$

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 equation (1) and (2), we get

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

∴ 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)

$\sf{\implies \dfrac{2}{3}BE+\dfrac{2}{3}AD > \dfrac{2}{3}CF}$

$\sf{\implies BE+AE>CF\;\;\;\;\;\;\;\bigg[Centroid\;G\;divides\;median\;in\;the\;ratio\;2:1 \bigg]}$

$\sf{\bigg[\therefore AG=\dfrac{2}{3}AD, BG=\dfrac{2}{3}BE\;and\;CG=\dfrac{2}{3}CF\bigg]}$

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

Hence Proved.