Question
In the given figure, $\angle\text{A}=\angle\text{CED},$ prove that $\triangle\text{CAB}\sim\triangle\text{CED}.$ Also, find the value of x.
✓
Answer
We have,
$\angle\text{A}=\angle\text{CED}$
In $\triangle\text{CAB}$ and $\triangle\text{CED}$
$\angle\text{A}=\angle\text{CED}$
$\angle\text{C}=\angle\text{C}$
$\triangle\text{CAB}\sim\triangle\text{CED}$ (By AA criteria)
Now, $\frac{\text{CA}}{\text{CE}}=\frac{\text{AB}}{\text{ED}}$
$\Rightarrow\frac{7+8}{10}=\frac{9}{\text{x}}$
$\Rightarrow\text{x}=\frac{9\times10}{15}$
$\Rightarrow\text{x}=\frac{90}{15}$
$\Rightarrow\text{x}=6\text{cm}.$
$\angle\text{A}=\angle\text{CED}$
In $\triangle\text{CAB}$ and $\triangle\text{CED}$
$\angle\text{A}=\angle\text{CED}$
$\angle\text{C}=\angle\text{C}$
$\triangle\text{CAB}\sim\triangle\text{CED}$ (By AA criteria)
Now, $\frac{\text{CA}}{\text{CE}}=\frac{\text{AB}}{\text{ED}}$
$\Rightarrow\frac{7+8}{10}=\frac{9}{\text{x}}$
$\Rightarrow\text{x}=\frac{9\times10}{15}$
$\Rightarrow\text{x}=\frac{90}{15}$
$\Rightarrow\text{x}=6\text{cm}.$
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