Question

In the given figure, if $AB = AC$, then prove that $BE = EC$. 

Hint:

In a triangle with an incircle, tangents from the same external point are always equal. Use this property to establish equal segments.

Answer 1

Step 1: Given. 
A circle is inscribed in $\triangle ABC$ touching sides $BC, CA, AB$ at $E, F, D$ respectively, and $AB = AC$. 
Step 2: Use tangent properties. 
Tangents drawn from an external point to a circle are equal in length. Thus, 

Step 3: Since $AB = AC$. 
\[ AD + BD = AF + CF \] Substitute $AD = AF$: \[ BD = CF \] 
Step 4: Add equal parts. 
\[ BD + BE = CF + CE \Rightarrow BE = CE \] 
Step 5: Conclusion. 
Hence, in $\triangle ABC$ where $AB = AC$, we have $\boxed{BE = EC}$.