Question

In the figure, ABC and ABD are two triangles on the 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 : in the figure, $△ABCand△ABD$ are on he same base AB and line CD is bisected by AB at O. Hence, CO = OD.
To prove : $ar\left(△ABC\right)=ar\left(△ABD\right)$
Construction : Draw $CL\perp ABandDM\perp AB$


Proof : In $△CLOand△EMO\angle L=\angle MCO=OD\angle COL=\angle DOM\therefore △CLO\cong △DMO$
$\therefore CL=DM$ (By CPCT)
Now ar $\left(△ABC\right)=\frac{1}{2}Base\times Height=\frac{1}{2}\times AB\times CLandar\left(△ADB\right)=\frac{1}{2}\times AB\times DM=\frac{1}{2}AB\times CL$