Question

In the adjoining figure, if \(\dfrac{AD}{BD} = \dfrac{AE}{EC}\) and \(\angle BDE = \angle CED\), prove that \(\triangle ABC\) is an isosceles triangle.

Hint:

Use the Basic Proportionality Theorem and congruence rules for triangle equality.

Answer 1

Given:
\[ \frac{AD}{BD} = \frac{AE}{EC} \quad \text{and} \quad \angle BDE = \angle CED \]
To prove: \(\triangle ABC\) is isosceles.

Step 1: Consider triangles \(\triangle BDE\) and \(\triangle CED\)
- Given \(\angle BDE = \angle CED\) (Given)
- \(DE\) is common side.
- Also, \(\frac{AD}{BD} = \frac{AE}{EC}\) (Given)

Step 2: Use Side-Angle-Side similarity
In \(\triangle BDE\) and \(\triangle CED\), by SAS similarity:
\[ \triangle BDE \sim \triangle CED \]

Step 3: Corresponding angles of similar triangles are equal
\[ \Rightarrow \angle DBE = \angle ECD \]

Step 4: Show that \(\triangle ABC\) is isosceles
Note that \(\angle DBE\) and \(\angle ECD\) are parts of \(\angle ABC\) and \(\angle ACB\) respectively.
Since \(\angle DBE = \angle ECD\) and other corresponding parts are equal, it follows:
\[ \angle ABC = \angle ACB \]
Hence, \(\triangle ABC\) has two equal angles and therefore two equal sides.
So, \(\triangle ABC\) is isosceles.

Final Statement:
\[ \triangle ABC \text{ is an isosceles triangle.} \]