Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
MediumProve that the line segment joining the midpoints of two opposite sides of a parallelogram divides the parallelogram into two parallelograms with equal area.
(N/A) Let $ABCD$ be a parallelogram where $E$ and $F$ are the midpoints of sides $AB$ and $CD$ respectively.
Since $AB \parallel CD$ and $AB = CD$,it follows that $AE = EB = \frac{1}{2} AB$ and $CF = FD = \frac{1}{2} CD$.
Thus,$AE = FC$ and $AE \parallel FC$,which implies that $AEFC$ is a parallelogram.
Similarly,$EBFD$ is a parallelogram because $EB = FD$ and $EB \parallel FD$.
Since $ABCD$ is a parallelogram,the height $h$ between sides $AB$ and $CD$ is constant.
The area of parallelogram $AEFC = AE \times h$ and the area of parallelogram $EBFD = EB \times h$.
Since $AE = EB$,it follows that $\text{Area}(AEFC) = \text{Area}(EBFD)$.
Thus,the line segment $EF$ divides the parallelogram $ABCD$ into two parallelograms of equal area.
Since $AB \parallel CD$ and $AB = CD$,it follows that $AE = EB = \frac{1}{2} AB$ and $CF = FD = \frac{1}{2} CD$.
Thus,$AE = FC$ and $AE \parallel FC$,which implies that $AEFC$ is a parallelogram.
Similarly,$EBFD$ is a parallelogram because $EB = FD$ and $EB \parallel FD$.
Since $ABCD$ is a parallelogram,the height $h$ between sides $AB$ and $CD$ is constant.
The area of parallelogram $AEFC = AE \times h$ and the area of parallelogram $EBFD = EB \times h$.
Since $AE = EB$,it follows that $\text{Area}(AEFC) = \text{Area}(EBFD)$.
Thus,the line segment $EF$ divides the parallelogram $ABCD$ into two parallelograms of equal area.
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