Question

In a quadrilateral ABCD, M is the midpoint of AC. Prove that $\mathrm{ar}\left(\boxed{}ABMD\right)=\mathrm{ar}(\boxed{}DMBC)$.

Answer 1

Since, M is the mid-point of AC, AM is the median of $△$ADC.
We know that a median divides a triangle in two triangles of equal area.
Hence,ar($△$AMD) = ar($△$CMD) ......(i)
Also, BM is the median of $△$ABC.
Hence,ar($△$AMB) = ar($△$CMB).....(ii)
By adding equations (i) and (ii), we get:
ar($△$AMD)+ar($△$AMB) = ar($△$CMD) +ar($△$CMB)
$\Rightarrow \mathrm{ar}\left(\square ABMD\right)=\mathrm{ar}(\square DMBC)$