In the parallelogram ABCD, the side AB is produced to the point X, so that BX = AB. The line DX cuts BC at E. Prove that

(i) DBXC is a parallelogram.

(ii) Area (△ AED) = 2 x area (△ CEX).

(i) Given: ABCD is a parallelogram, the side AB is extended to the point X such that BX = AB.

The line DX cuts BC at E.

To prove: DBXC is a parallelogram.

Proof: Since ABCD is a parallelogram, we know that:

AB = DC and AB ∥ DC

Given that AB = BX, we get:

⇒ DC = BX and DC ∥ BX

Since one pair of opposite sides of quadrilateral DBXC is both equal and parallel, it follows that:

Hence, DBXC is a parallelogram.

(ii) Given: From part (i), DBXC is a parallelogram, which implies that E is the midpoint of parallelogram DBXC.

To prove: Area (△ AED) = 2 x area (△ CEX)

Proof: Since E is the midpoint of parallelogram DBXC, the line AE acts as a median in Δ ADX, dividing it into two equal areas:

ar.(Δ AED) = ar.(Δ AEX) ……………….(1)

Similarly, BE is a median in Δ AED, dividing Δ AEX into two equal triangles:

ar.(Δ AEB) = ar.(Δ BEX) = $\dfrac{1}{2}$ ar.(Δ AEX) ……………….(2)

Since XE is a median in Δ BXC, it divides it into two equal areas:

ar.(Δ CEX) = ar.(Δ BEX)

Substituting equation (2), we get:

ar.(Δ CEX) = $\dfrac{1}{2}$ ar.(Δ AEX)

Since from equation (1), ar.(Δ AEX) = ar.(Δ AED), we can write:

ar.(Δ CEX) = $\dfrac{1}{2}$ ar.(Δ AED)

So, ar.(Δ AED) = 2 x ar.(Δ CEX)

Hence, ar.(Δ AED) = 2 x ar.(Δ CEX).