Question
In figure, $ABC$ and $ABD$ are two triangles on the base $AB$. If line segment $CD$ is bisected by $AB$ at $O$, show that $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ABD}).$
✓
Answer
Given that $CD$ is bisected by $AB$ at $O$
To prove: $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ABD}).$
Construction: Draw $\text{CP}\perp\text{AB}$ and $\text{DQ}\perp\text{AB}.$
Proof: $\text{ar}(\triangle\text{ABC})=\frac{1}{2}\times\text{AB}\times\text{CP}\ ⋅⋅⋅⋅⋅ (1)$
$\text{ar}(\triangle\text{ABD})=\frac{1}{2}\times\text{AB}\times\text{DQ}\ ⋅⋅⋅⋅⋅ (2)$ In
$\triangle\text{CPO}$ and $\triangle\text{DQO}$ $\angle\text{CPO}=\angle\text{DQO}$ [Each $90^\circ$]
Given that, $CO = OD$ $\angle\text{CPO}=\angle\text{DQO}$ [Vertically opposite angles are equal]
Then, $\triangle\text{CPO}\cong\text{DQO}$ [By $AAS$ condition]
$\therefore$ $CP = DQ (3)$ $[C.P.C.T]$ Compare equation $(1), (2)$ and $(3)$
$\therefore\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ABD}).$
To prove: $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ABD}).$
Construction: Draw $\text{CP}\perp\text{AB}$ and $\text{DQ}\perp\text{AB}.$
Proof: $\text{ar}(\triangle\text{ABC})=\frac{1}{2}\times\text{AB}\times\text{CP}\ ⋅⋅⋅⋅⋅ (1)$
$\text{ar}(\triangle\text{ABD})=\frac{1}{2}\times\text{AB}\times\text{DQ}\ ⋅⋅⋅⋅⋅ (2)$ In
$\triangle\text{CPO}$ and $\triangle\text{DQO}$ $\angle\text{CPO}=\angle\text{DQO}$ [Each $90^\circ$]
Given that, $CO = OD$ $\angle\text{CPO}=\angle\text{DQO}$ [Vertically opposite angles are equal]
Then, $\triangle\text{CPO}\cong\text{DQO}$ [By $AAS$ condition]
$\therefore$ $CP = DQ (3)$ $[C.P.C.T]$ Compare equation $(1), (2)$ and $(3)$
$\therefore\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ABD}).$
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