Question

If each diagonal of a quadrilateral divides it into two triangles to equal areas then prove that quadrilateral is a parallelogram.

Answer 1

Proof:

With respect to diagonal $AC,$

$\Rightarrow$ $ar\left(△ABC\right)=ar\left(△ACD\right)\dots \left(i\right)$

$\Rightarrow$ $ar\left(△ABC\right)+ar\left(△ACD\right)=ar\left(ABCD\right)\dots \left(ii\right)$

From $\left(i\right)$ and $\left(ii\right),$

$\Rightarrow 2ar\left(△ABC\right)=ar\left(ABCD\right)\dots \left(iii\right)$

With respect to diagonal $BD,$

$\Rightarrow ar\left(△ABD\right)=ar\left(△BCD\right)\dots \left(iv\right)$

$\Rightarrow$ $ar\left(△ABD\right)+ar\left(△BCD\right)=ar\left(ABCD\right)\dots \left(v\right)$

From $\left(iv\right)$ and $\left(v\right),$

$\Rightarrow 2ar\left(△ABD\right)=ar\left(ABCD\right)\dots \left(vi\right)$

From $\left(iii\right)$ and $\left(vi\right)$ we get,

$\Rightarrow 2ar\left(△ABC\right)=2ar\left(△ABD\right)$

$\Rightarrow ar\left(△ABC\right)=ar\left(△ABD\right)$

Since $△ABC$ and $△ABD$ are on the same base $AB$ and have equal area. Therefore, they must have equal corresponding altitudes.

i.e. Altitude from $C$ of $△ABC=$ Altitude from $D$ of $△ABD$

$\Rightarrow$ $DC\parallel AB$

Similarly, we can prove that $AD\parallel BC.$

So, opposite sides of quadrilateral $ABCD$ are parallel which implies that $ABCD$ is a parallelogram.