In ∆ABC; D, E and F are mid-points of sides BC,AC and AB, respectively. A line through C drawn parallel to DE meets FE produced to G. Show that ar (∆FDE) = ar (∆EGC)

∵D and F are the mid-points of BC and AB respectively. Using mid-point theorem in ∆ABC

i.e. FD = AE and FD||AE i.e. AFDE is a parallelogram.

∴ ar (∆FAC) = ar (∆FDE) ,…(i)

[∵ FE diagonal of ||Parallelogram which divides two equal triangles]

In ∆FAE and ∆GCE,

∠1 = ∠2 [alternate interior angles]

AE = EC [given]

∠3 = ∠4 [vertically opposite angles]

∴ ∆FAE =∆GCE [by ASA Congruence rule]

So, ar (∆FAE) = ar (∆GCE)

i. e. ar(∆FAE)=ar(∆EGC) …(ii)

from Eqs. (i) and (ii), we have

ar (∆FDE) = ar (∆EGC)