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**Hint:**We are given that we have a $\left( \vartriangle ABC \right)$ and a parallelogram that lie on the same base we have to show that

$\left( \vartriangle ABC \right)=\dfrac{1}{2}$ar (||gm ABEP)

To do so we do construction we draw BH||AC that meet FE at H

Then we use that diagonal of parallelogram to divide it into two congruent triangles and we use that two parallelograms on the same base between the same parallel are equal in area. Using them we get our required solution.

**Complete step by step answer:**

We are given that $\left( \vartriangle ABC \right)=\dfrac{1}{2}$ar (||gm ABEP) are on the same base while AC is between the same parallel AB and EF

So, using these data we will first draw our figure to get better understanding of problem

Now as Ab is common to $\vartriangle ABC$and parallelogram ABEF, so, common base in AB. So, we have,

Now to solve this problem and get our required solution,

Through B we draw BH||AC to meet FE provided at H

Such that ABHC is a parallelogram.

As we know that diagonal of parallelogram divide it into two congruent,

So, as CB is diagonal at ||ABHC

So, $\vartriangle ABC\cong \vartriangle BHC$

A congruent triangle has some area.

So, $ar\left( \vartriangle ABC \right)=ar\left( \vartriangle BHC \right)..........(i)$

Now as

$ar\left( ||gmABHC \right)=ar\left( \vartriangle ABC \right)+ar\left( BHC \right)$

Why (i), we get

$ar\left( ||gmABHC \right)=ar\left( \vartriangle ABC \right)+ar\left( BHC \right)$

So, we get

$ar\left( \vartriangle ABC \right)=\dfrac{1}{2}\left( ar\left( ||gm\,ABHC \right) \right).......(ii)$

Now

We know that two parallelograms on same base and between same parallel are equal in equal we have that

||gm ABHC and ||gm ABEP are on the same base AB and between the same parallel AB and EF.

So,

$ar\left( ||gmABHC \right)=ae\left( ||gmABEF \right)........(iii)$

Now, using (iii) and (ii)

$ar\left( \vartriangle ABC \right)=\dfrac{1}{2}\left( ||gmABEF \right)$

Hence proved

**Note:**

Parallelogram divide into two parts by the diagonal as we can see that through diagonal we have $\vartriangle ABC\,\operatorname{and}\vartriangle ACD$ in ||gm ABCD.

In $\vartriangle ABC\,\operatorname{and}\vartriangle ACD$, AB=CD (opposite sides are equal in parallelogram)

BC=AD (opposite sides are equal common)

AC=AC

So, using SSS rule, we have these two are congruent.

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