Question
In the given figure, $\angle1=\angle2$ and $\frac{\text{AC}}{\text{BD}}=\frac{\text{CB}}{\text{CE}}$
Prove that $\triangle\text{ACB}\sim\triangle\text{DCE}.$
Prove that $\triangle\text{ACB}\sim\triangle\text{DCE}.$
✓
Answer
$\angle1=\angle2$ (given)
$\frac{\text{AC}}{\text{BD}}=\frac{\text{CB}}{\text{CE}}\Rightarrow\frac{\text{AC}}{\text{CB}}=\frac{\text{BD}}{\text{CE}}$ (given)
Also, $\angle2=\angle1$ $\big[\therefore\text{BD}=\text{DC}\big]$
Thus, $\frac{\text{AC}}{\text{CB}}=\frac{\text{BD}}{\text{CE}}$
and $\angle2=\angle1$
Therefore, by SAS similarity criterion $\triangle\text{ACB}\sim\triangle\text{DCE}$
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