Question

The given figure shows a pentagon $ABCDE$ .$EG$ drawn parallel to $DA$ meets $BA$ producedat G and $CF$ drawn parallel to $DB$ meets $AB$ produced at $F$ .
Prove that the area of pentagon $ABCDE$ is equal to the area of triangle $GDF$ .

Answer 1

Consider $△DGA$ and $△AED$

We know that both the triangles have the same base $AD$ and lie between the parallel lines $AD$ and $EG$.

So we get

Area of $△DGA$$=$ Area of $△AED$.....(1)

Consider $△DBC$ and $△BFD$

We know that both the triangles have the same base $DB$ and lie between the parallel lines $BD$ and $CF$.

So we get

Area of $△DBF$ $=$ Area of $△BCD$......(2)

By adding both the questions

Area of $△DGA$ $+$ Area of $△DBF$$=$ Area of $△AED$ $+$Area of $△BCD$

By adding $△ABD$ both sides

Area of $△DGA$$+$ Area of $△DBF$ $+$ Area of $△ABD$$=$Area of $△AED$$+$Area of $△BCD$

By adding $△ABD$ both sides

Area of $△DGA$ $+$ Area of $△DBF$ $+$ Area of $△ABD$ $=$ Area of $△AED$$+$Area of $△BCD$$+$ Area of $△ABD$

So we get

Area of $△DGF$ $=$ Area of pentagon $ABCDE$

Therefore, it is proved that $ar\left(pentagonABCDE\right)$=$ar\left(△DGF\right)$.