Question

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

Answer 1

Since $CE$ perpendicular to $AB$ and $DF$ perpendicular to $AB$

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

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

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

ar$\left(△ABD\right)=\frac{1}{2}\times$ base$\times$

height$=\frac{1}{2}\times AB\times DF$

Since ar$\left(△ABC\right)=$ar$\left(△ABD\right)$ (given)

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

Now, in $CDEF$,

$CE\parallel DF$ and $CE=DF$

Since one pair of opposite sides are equal and parallel.

Hence, $CDFE$ is a parallelogram.

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

$\therefore CD\parallel AB$

Hence proved.