Q5 of 24 Page 296

Let us prove that two medians of triangle are together greater than the third median.


Given: AD, BE and FC are medians.


Extend AD to H such that (AG = HG) then join BH and CH.


In Δ ABH,


F is mid-point and G is centroid, so use mid-point theorem,


So, FG || BH


Similarly, GC || BH ….(1)


And BG || HC …..(2)


From (1) and (2),


We get,


BGCH is a parallelogram,


So, BH = GC …..(3) (opp. Sides of a parallelogram are equal)


In Δ BGH,


BG + GH > BH (sum of two sides greater than third side)


⇒ BG + AG > GC


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


Hence proved.


More from this chapter

All 24 →
3

Let us prove that in an equilateral triangle, circumcentre, incentre, centriod, orthocenter will coincide.

 4

AD, BE and CF are three medians of a triangle ABC. Let us prove that the centriod of ABC and DEF are the same point.

 6

AD, BE and CF are three medians of ΔABC. Let us prove that

(i) 4 (AD + BE + CF) > 3 (AB + BC + CA)


(ii) 3(AB + BC + CA) > 2(AD + BE + CF)

 7

Three medians AD, BE and CF of ΔABC intersect each other at the point G. If area of ΔABC is 36 sq. cm, let us calculate.

(i) Area of ΔAGB


(ii) Area of ΔCGE


(iii) Area of quadrilateral BDGF.