Q5 of 24 Page 256

In the figure D, E are points on the sides AB and AC respectively of ∆ABC such that ar(∆DBC) = ar(∆EBC). Prove that DE || BC.


Consider h1 and h2 as heights of ∆DBC and ∆EBC from points D and E respectively


Given area(∆DBC) = area(∆EBC)


The base of both the triangles is common i.e. BC


Height of ∆DBC = h1


Height of ∆EBC = h2


⇒ × BC × h1 = × BC × h2


⇒ h1 = h2


Which means points D and E are on the same height from segment BC which implies that line passing through both the points i.e. D and E is parallel to the BC


Therefore DE || BC


More from this chapter

All 24 →
3

In the figure, ∆ABC and ∆ABD are two triangles on the same base AB. If line segment CD is bisected by at O, show that ar(∆ABC) = ar(∆ABD)

 4

In the figure, ∆ABC, D, E, F are the midpoints of sides BC, CA and AB respectively. Show that

(i) BDEF is a parallelogram


(ii) ar(∆DEF) = ar(∆ABC)


(iii) ar(BDEF) = ar(∆ABC)


 6

In the figure, XY is a line parallel to BC is drawn through A. If BE || CA and CF || BA are drawn to meet XY at E and F respectively. Show that ar(∆ABE) = ar (∆ACF).

 7

In the figure, diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar(∆AOD) = ar(∆BOC).