| In figure, if \[\frac{AD}{DC}=\frac{BE}{EC}\] and \[\angle CDE=\angle CED,\] then: |
 |
A) \[\text{BC}=\text{AC}\]
B) \[\text{AB}=\text{AC}\]
C) \[\text{AB}=B\text{C}\]
D) \[CE=DE\]
Correct Answer: A
Solution :
| [a] In \[\Delta ABC,\] |
| we have \[\frac{AD}{DC}=\frac{BE}{EC}\] (Given) |
| Therefore, by the converse of basic proportionality theorem. |
| we have. \[DE||AB\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\angle CDE=\angle CAB\] |
| and \[\angle CED=\angle CBA\] (Corresponding angles) |
| But, \[\angle CDE=\angle CED\] (Given) |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\angle CAB=\angle CBA\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\angle A=\angle B\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,BC=AC\] |
| (Sides opposite to equal angles are equal) |
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