Question

In given figure, if ${\displaystyle \frac{AD}{DC}}={\displaystyle \frac{BE}{EC}}$ and $\mathrm{\angle }CDE=\mathrm{\angle }CED$, prove that $△CAB$ is isosceles.
                        

Answer 1

Hint: To solve this question, we will use the concept of converse of basic proportionality theorem. This states that if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Complete step-by-step answer:
Given that,
In $△ABC$,
 $\Rightarrow {\displaystyle \frac{AD}{DC}}={\displaystyle \frac{BE}{EC}}$,
Therefore, by using the converse of basic proportionality theorem,
We have,
$\Rightarrow DE\parallel AB$
And
$\Rightarrow \mathrm{\angle }CDE=\mathrm{\angle }CAB$ and [corresponding angles]
$\Rightarrow \mathrm{\angle }CED=\mathrm{\angle }CBA$
And we have given,
$\Rightarrow \mathrm{\angle }CDE=\mathrm{\angle }CED$ [given]
So,
$\Rightarrow \mathrm{\angle }CBA=\mathrm{\angle }CAB$
Or we can say that,
$\Rightarrow \mathrm{\angle }B=\mathrm{\angle }A$
We know that the sides that are opposite to equal angles are equal.
Therefore,
$\Rightarrow BC=AC$
Hence,
$△CAB$ is isosceles.

Note: Whenever we ask such types of questions, we have to remember the converse of basic proportionality theorem. On the other hand, the basic proportionality theorem states that if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio. It is also known as Thales theorem.