Q6 of 24 Page 256

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).

Given XY || BC and BE || CA and CF || BA


XY || BC implies EA || BC and AF || BC as points E, A and F lie on XY line


Consider quadrilateral ACBE


AC || EB and EA || BC opposite sides are parallel


Therefore, quadrilateral ACBE is a parallelogram with AB as the diagonal


the diagonal divides the area of parallelogram in two equal parts


⇒ area(ΔABE) = area(ΔABC) …(i)


Consider quadrilateral ABCF


AB || FC and AF || BC opposite sides are parallel


Therefore, quadrilateral ABCF is a parallelogram with AC as the diagonal


the diagonal divides the area of parallelogram in two equal parts


⇒ area(ΔACF) = area(ΔABC) …(ii)


From (i) and (ii)


area(∆ABE) = area(∆ACF)


More from this chapter

All 24 →
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)


 5

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.

 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).

 8

In the figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

(i) ar (∆ACB) = ar (∆ACF)


(ii) ar (AEDF) = ar (ABCDE)