Question

$ABC$ and $ABD$ are two triangles on the same base $AB.$ If line-segment $CD$ is bisected by $AB$ at $O,$ show that $ar\left(ABC\right)=ar\left(ABD\right).$

Answer 1

Given:$△ABC$ and $△ABD$ on the same base $AB$ and $AB$ bisects $CD$

$OC=OD$ .....$\left(1\right)$

To prove:ar$\left(△ABC\right)=$ar$\left(△ABD\right)$

Proof:

In $△ACD,$

Since $OC=OD$ from $\left(1\right)$

$\therefore ,OA$ is the median

$\Rightarrow$ ar$\left(△AOC\right)=$ar$\left(△AOD\right)$ ....$\left(2\right)$

Since median divides the triangle into equal areas

Similarly, in $△BCD$

since $OC=OD$ from $\left(1\right)$

$\therefore OB$ is the median

$\Rightarrow$ ar$\left(△BOC\right)=$ar$\left(△BOD\right)$ ....$\left(3\right)$

Since median divides the triangle into equal areas

Adding eqns$\left(2\right)$ and $\left(3\right)$ we get

ar$\left(△AOC\right)+$ar$\left(△BOC\right)=$ar$\left(△AOD\right)+$ar$\left(△BOD\right)$

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

Hence proved.