Question

Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.

Answer 1

Given:$△ABC$ and $△ABD$ are two trianles on same base $AB$ such that area$\left(△ABC\right)=$area$\left(△ABD\right)$.

Join $CD$

To prove:$CD\parallel AB$

Construction:Draw the altitutdes of $△ABC$ and $△ABD$ throught $C$ and $D$ and meeting base at $E$ and $F$ respectively.

Proof:Since $CE\perp AB$ and $DF\perp AB$(by construction)

Since, lines perpendicular to same line are parallel to each other.

$\therefore CE\parallel DF$ .......$\left(1\right)$

Now, area$\left(△ABC\right)=\frac{1}{2}\times AB\times CE$

area$\left(△ABD\right)=\frac{1}{2}\times AB\times DF$

Since area$\left(△ABC\right)=$area$\left(△ABD\right)$(Given)

$\therefore CF=DF$ .........$\left(2\right)$

Now, In $CDFE,CF\parallel DF$ and $CE\parallel DF$

Since one pair of opposite sides are equal and parallel

Hence $CDFE$ is a parallelogram.

So, $CD\parallel EF$ (opposite sides of parallelogram are parallel)

$\therefore CD\parallel AB$

Hence, proved.