BD is one of the diagonals of a quad.

Question:

$B D$ is one of the diagonals of a quad. $A B C D$. If $A L \perp B D$ and $C M \perp B D$, show that ar (quad. $A B C D$ ) $=\frac{1}{2} \times B D \times(A L+C M)$.

Solution:

ar(quad. ABCD) = ar(∆​ABD) + ar (∆DBC)

$\operatorname{ar}(\Delta A B D)=\frac{1}{2} \times$ base $\times$ height $=\frac{1}{2} \times B D \times A L$     ...(i)

$\operatorname{ar}(\Delta D B C)=\frac{1}{2} \times B D \times C L$                      ...(ii)

From (i) and (ii), we get:

$\operatorname{ar}($ quad $A B C D)=\frac{1}{2} \times B D \times A L+\frac{1}{2} \times B D \times C L$

$\Rightarrow \operatorname{ar}($ quad $A B C D)=\frac{1}{2} \times B D \times(A L+C L)$

Hence, proved.