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in the adjoining figure, ABCD is a parallelogram and BF = EF. Prove that Area of EFG = 1/4 Area of ABCD.​

in the adjoining figure, ABCD is a parallelogram and BF = EF. Prove that Area of EFG-example-1

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Answer:

Step-by-step explanation: Let's height of the parallelogram ABCD as h

height h can be determined by drawing a perpendicular line from one of the vertices of the base.

Now, let's consider triangle ABE and triangle CDF. They have the same base, AB = CD, and the same height, h. Therefore, their areas are equal:

Area(AEB) = Area(CDF) -----------(1)

Since ABCD is a parallelogram, AD || BC. This implies that angle BCD = angle ADB.

angle BCD = angle ADB

angle CDE = angle ADE

CDE is congruent to triangle ADE by the angle-side-angle (ASA) congruence criterion

CE = AE.

given that BF = EF

means that BE = EF + BF = EF + EF = 2EF

Area(CDE) = A - 2x -----------(2)

From equation (1), we have Area(AEB) = Area(CDF), which can be expressed as:

h * AB = h * CD

Since AB = CD, we can cancel the height h, resulting in:

AB = CD

Now, consider the segment DE. We know that DE = AB - AE = AB - CE.

Since AB = CD, we can rewrite DE as:

DE = CD - CE

Substituting the expression for DE in equation (2), we get:

Area(CDE) = A - 2x = Area(CDF) = h * DE = h * (CD - CE)

Substituting the value of CD - CE from the above equation, we have:

A - 2x = h * (CD - CE) = h * DE

Since BF = EF, we know that DE = 2EF.

Substituting this value, we get:

A - 2x = h * 2EF

Dividing both sides by 2, we have:

(A - 2x) / 2 = h * EF

Simplifying further, we get:

(A - 2x) / 2h = EF

But EF = BF, so:

(A - 2x) / 2h = BF

Multiplying both sides by h, we obtain:

(A - 2x) / 2 = h * BF

Substituting the value of h * BF with the area of triangle ABCD (A), we get:

(A - 2x) / 2 = A / 4

Cross-multiplying, we have:

2(A - 2x) = A

Expanding the equation, we get:

2A - 4x = A

Subtracting 2A from both sides, we get:

-4x = -A

Dividing both sides by -4, we have:

x = A / 4

Therefore, the area of triangle EFG (x) is equal to one-fourth the area of parallelogram ABCD (A).

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