Question

ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:

(i) ar (ADEG) = ar (GBCE)

(ii) ar (Δ EGB) = $\frac{1}{6}$ ar (ABCD)

(iii) ar (Δ EFC) = $\frac{1}{2}$ ar (Δ EBF)

(iv) ar (Δ EBG) = ar (Δ EFC)

(v) Find what portion of the area of parallelogram is the area of Δ EFG

Answer 1

Given:

ABCD is a parallelogram

G is a point such that AG = 2GB

E is a point such that CE = 2DE

F is a point such that BF = 2FC

To prove:

(i)

(ii)

(iii)

(iv)

What portion of the area of parallelogram ABCD is the area of ΔEFG

Construction: draw a parallel line to AB through point F and a perpendicular line to AB through

PROOF:

(i) Since ABCD is a parallelogram,

So AB = CD and AD = BC

Consider the two trapeziums ADEG and GBCE:

Since AB = DC, EC = 2DE, AG = 2GB

, and

, and

So, and

Since the two trapeziums ADEG and GBCE have same height and their sum of two parallel sides are equal

Since

So

Hence

(ii) Since we know from above that

. So

Hence

(iii) Since height of triangle EFC and triangle EBF are equal. So

Hence

(iv) Consider the trapezium in which

(From (iii))

Now from (ii) part we have

(v) In the figure it is given that FB = 2CF. Let CF = x and FB = 2x

Now consider the tow triangles CFI and CBH which are similar triangles

So by the property of similar triangle CI = k and IH = 2k

Now consider the triangle EGF in which

Now

(Multiply both sides by 2)

…… (2)

From (1) and (2) we have