1. **Problem statement:** Given an isosceles triangle $BCD$ with vertex at $B$, points $E$ on $BC$ and $F$ on the ray opposite to $DB$ such that $CE = DF$. Lines $EH$ and $FG$ are perpendicular to $DC$ with $H, G \in DC$. $I$ is the intersection of $EF$ and $CD$. Given $EI = 15$ cm, find the length of $EF$. 2. **Key properties and setup:** - Since $BCD$ is isosceles at $B$, $BC = BD$. - $CE = DF$ means the segments from $C$ to $E$ and from $D$ to $F$ are equal. - $EH \perp DC$ and $FG \perp DC$ imply $EH$ and $FG$ are heights from $E$ and $F$ to $DC$. - $I$ lies on $CD$ and on $EF$. 3. **Geometric insight:** - Because $CE = DF$ and $E$ and $F$ are constructed symmetrically relative to $B$, $EF$ is parallel to $CD$. - Points $H$ and $G$ are projections of $E$ and $F$ onto $DC$, so $EH$ and $FG$ are heights. 4. **Using the intercept theorem and symmetry:** - Since $EF$ is parallel to $CD$, $I$ is the midpoint of $EF$. - Given $EI = 15$ cm, and $I$ is midpoint, then $EF = 2 \times EI = 2 \times 15 = 30$ cm. 5. **Final answer:** $$EF = 30 \text{ cm}$$