Question: Given: $AB \cong BC$, $DE$ bisects $AB$ and $DE$ bisects $BC$. Prove: $BD \cong BE$. Note: quadrilateral properties are not permitted in this proof.

1. **State the problem:** We are given that $AB \cong BC$, and that $DE$ bisects both $AB$ and $BC$. We need to prove that $BD \cong BE$. 2. **Recall definitions and properties:** - A segment bisector divides a segment into two equal parts. - Given $DE$ bisects $AB$, then $AD \cong DB$. - Given $DE$ bisects $BC$, then $BE \cong EC$. - Given $AB \cong BC$ means the lengths of $AB$ and $BC$ are equal. 3. **Write down what is known:** - $AB \cong BC$ (Given) - $AD \cong DB$ (Since $DE$ bisects $AB$) - $BE \cong EC$ (Since $DE$ bisects $BC$) 4. **Express $AB$ and $BC$ in terms of segments:** Since $D$ lies on $AB$, $AB = AD + DB$. Since $E$ lies on $BC$, $BC = BE + EC$. 5. **Use the bisector equalities:** Since $AD \cong DB$, $AD = DB$. Since $BE \cong EC$, $BE = EC$. 6. **Substitute into $AB \cong BC$:** $$AB = AD + DB = DB + DB = 2DB$$ $$BC = BE + EC = BE + BE = 2BE$$ Since $AB \cong BC$, lengths are equal: $$2DB = 2BE$$ 7. **Simplify the equation:** Divide both sides by 2: $$\cancel{2}DB = \cancel{2}BE$$ $$DB = BE$$ 8. **Conclusion:** Since $DB = BE$, segments $BD$ and $BE$ are congruent. **Therefore, $BD \cong BE$ is proven.**