In figure below, $ \mathrm{ABC} $ and $ \mathrm{ABD} $ are two triangles on the same base $ \mathrm{AB} $. If line- segment $ \mathrm{CD} $ is bisected by $ \mathrm{AB} $ at $ \mathrm{O} $, show that $ \operatorname{ar}(\mathrm{ABC})=\operatorname{ar}(\mathrm{ABD}) $.
"

Given:

$ABC$ and $ABD$ are two triangles on the base $AB$.

Line segment $CD$ is bisected by $AB$ at $O$.

To do:

We have to show that $ar(\triangle ABC) = ar(\triangle ABC)$.

Solution:

$CO = OD$

Draw $CL \perp AB$ and $DM \perp AB$

In $\Delta CLO$ and $\triangle D M O$,

$\angle \mathrm{L}=\angle \mathrm{M}$

$\mathrm{CO}=\mathrm{OD}$

$\angle \mathrm{COL}=\angle \mathrm{DOM}$           (Vertically opposite angles)

Therefore, by AAS axiom,

$\Delta CLO \cong \triangle D M O$

This implies,

$\mathrm{CL}=\mathrm{DM}$             (CPCT)

$ar(\triangle \mathrm{ABC})=\frac{1}{2}\times$ Base $\times$ Height

$=\frac{1}{2} \times \mathrm{AB} \times \mathrm{CL}$.........(i)

$ar(\triangle \mathrm{ADB})=\frac{1}{2} \times \mathrm{AB} \times \mathrm{DM}$

$=\frac{1}{2} \mathrm{AB} \times \mathrm{CL}$...........(ii)                      (Since $\mathrm{DM}=\mathrm{CL}$)

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

$ar(\triangle \mathrm{ABC})=ar(\triangle \mathrm{ABD})$

Hence proved.

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Updated on: 10-Oct-2022

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