Question
If $\text{AE = AD}$ and $\text{BD = CE}$. Prove that $\triangle AEB \cong \triangle ADC$
✓
Answer
We have,
$\text{AE = AD} [$ GIVEN $] \ldots(1)$ and $\text{CE = BD}[$ GIVEN $] \ldots(2)$
$\Rightarrow \text{AE + CE = AD + BD} [$ adding equation $(1) (2)]$
$\Rightarrow \text{AC = AB} \ldots(3)$
Now, in $\triangle AEB$ and $\triangle ADC$,
$\text{AE = AD}[$ given $]$
$\angle EAB =\angle DAC [$ common $] $
$AB = AC [$ from $(3)]$
$\triangle AEB \cong \triangle ADC [$by $\text{SAS}]$
$\text{AE = AD} [$ GIVEN $] \ldots(1)$ and $\text{CE = BD}[$ GIVEN $] \ldots(2)$
$\Rightarrow \text{AE + CE = AD + BD} [$ adding equation $(1) (2)]$
$\Rightarrow \text{AC = AB} \ldots(3)$
Now, in $\triangle AEB$ and $\triangle ADC$,
$\text{AE = AD}[$ given $]$
$\angle EAB =\angle DAC [$ common $] $
$AB = AC [$ from $(3)]$
$\triangle AEB \cong \triangle ADC [$by $\text{SAS}]$
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