MCQ
In the given figure, if $\angle\text{ADE}=\angle\text{ABC},$ then $\text{CE =}$
- A$2$
- B$5$
- ✓$\frac{9}{2}$
- D$3$
✓
Answer
Correct option: C.
$\frac{9}{2}$
Given: $\angle\text{ADE}=\angle\text{ABC}$
To find: The value of $\text{CE}$
Since $\angle\text{ADE}=\angle\text{ABC}$
$\therefore\text{DE}\|\text{BC} ($Two lines are parallel if the corresponding angles formed are equal$)$
According to basic proportionality theorem if a line is parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
In $\triangle\text{ABC},\ \text{DE}\|\text{BC}$
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\frac{2}{3}=\frac{3}{\text{EC}}$
$\text{EC}=\frac{3\times3}{2 }$
$\text{EC}=\frac{9}{2}$
Hence we got the result $C.$
To find: The value of $\text{CE}$
Since $\angle\text{ADE}=\angle\text{ABC}$
$\therefore\text{DE}\|\text{BC} ($Two lines are parallel if the corresponding angles formed are equal$)$
According to basic proportionality theorem if a line is parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
In $\triangle\text{ABC},\ \text{DE}\|\text{BC}$
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\frac{2}{3}=\frac{3}{\text{EC}}$
$\text{EC}=\frac{3\times3}{2 }$
$\text{EC}=\frac{9}{2}$
Hence we got the result $C.$
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