1. **Problem Statement:** We need to find the measure of angle $\angle BCD$ in the given geometric figure. 2. **Given Information:** - $\angle BEC = 65^\circ$ - $EB = ED$ (segments are equal) - $\angle EBD = 90^\circ$ and $\angle EDB = 90^\circ$ (right angles) 3. **Key Concept:** Since $EB = ED$ and both angles at $B$ and $D$ are right angles, triangle $EBD$ is isosceles right triangle with $EB = ED$ and $\angle EBD = \angle EDB = 90^\circ$. 4. **Step-by-step Solution:** - Since $EB = ED$ and $\angle EBD = \angle EDB = 90^\circ$, triangle $EBD$ is right isosceles. - $\angle BED$ is the remaining angle in triangle $EBD$, so $$\angle BED = 180^\circ - 90^\circ - 90^\circ = 0^\circ$$ which is impossible, so the right angles must be at $B$ and $D$ but not both in the same triangle. Instead, the problem likely means $\angle EBD = 90^\circ$ and $\angle EDB = 90^\circ$ separately at points $B$ and $D$ in different triangles. - Given $\angle BEC = 65^\circ$ and $EB = ED$, triangle $BEC$ and triangle $DEC$ share side $EC$. - Since $EB = ED$, triangle $EBD$ is isosceles with $EB = ED$. - $\angle BEC = 65^\circ$ is an exterior angle to triangle $BCD$ at vertex $C$. - By the exterior angle theorem, $\angle BEC = \angle BCD + \angle CBD$. - Since $\angle CBD = 90^\circ$ (given), then $$\angle BCD = \angle BEC - \angle CBD = 65^\circ - 90^\circ = -25^\circ$$ which is impossible. - Reconsidering, since $\angle CBD$ is right angle, $\angle BCD$ and $\angle BEC$ are complementary. - Therefore, $\angle BCD = 90^\circ - 65^\circ = 25^\circ$. 5. **Final Answer:** $$m \angle BCD = 25^\circ$$