In quadrilateral $ABCD$,$AM$ and $CN$ are altitudes on diagonal $BD$ drawn from $A$ and $C$ respectively. Prove that,$\operatorname{ar}(ABCD) = \frac{1}{2} \times BD \times (AM + CN)$.

(N/A) The area of quadrilateral $ABCD$ can be divided into the sum of the areas of two triangles,$\triangle ABD$ and $\triangle BCD$,by the diagonal $BD$.
$\operatorname{ar}(ABCD) = \operatorname{ar}(\triangle ABD) + \operatorname{ar}(\triangle BCD)$.
The area of a triangle is given by the formula $\frac{1}{2} \times \text{base} \times \text{height}$.
For $\triangle ABD$,the base is $BD$ and the height is $AM$. Thus,$\operatorname{ar}(\triangle ABD) = \frac{1}{2} \times BD \times AM$.
For $\triangle BCD$,the base is $BD$ and the height is $CN$. Thus,$\operatorname{ar}(\triangle BCD) = \frac{1}{2} \times BD \times CN$.
Substituting these into the first equation:
$\operatorname{ar}(ABCD) = (\frac{1}{2} \times BD \times AM) + (\frac{1}{2} \times BD \times CN)$.
Factoring out the common term $\frac{1}{2} \times BD$:
$\operatorname{ar}(ABCD) = \frac{1}{2} \times BD \times (AM + CN)$.
Hence,the area of the quadrilateral is proved.