Question

ABC is a triangle and DE is drawn parallel to BC, cutting the other sides at D and E. Join BE and CD. Prove that
1) area of triangle DBC =area of triangle EBC
2) area of triangle BDE = area of triangle CDE

Answer 1

triangles on the same base and between same parallel lines have equal areas
in case of triangles DBC and EBC.... BC is the same base and DE is parallel to BC
in case of triangles BDE and CDE... DE is the same base and DE is parallel to BC

Answer 2

Given,
ABC is a triangle
DE||BC
R.T.P.: 1) Area of DBC=Area of EBC
            2)area of BDE = area of CDE

soln.
by the theorem on area ,13.4, which states that area of a triangle is equal to 1/2 the area of a ||gm if it lies on the same base and between the same parallels of lines as the ||gm
here,
^ BDC & ^ BEC lie on the same base BC of the ||gm and between the same parallels
=> ^BDC=1/2 area of  ||gm DBCE    ...(i)
=> ^CEB=1/2 area of  ||gm DBCE    ...(ii)
from (i) & (ii),
^BDC= ^CEB (areas)
Hence,proved.

Similarly,
area of ^BDE = area of ^CDE (=1/2 area of  ||gm DBCE )
as they lie on the same base DE and between the same parallels
Hence,proved.