1. **State the problem:** Given a triangle ABC with $AB \perp AC$, $DE \perp FG$, $CD \cong CE$, and $m\angle B = 44^\circ$, find $m\angle FGB$. 2. **Given information:** - $m\angle B = 44^\circ$ (Given) - $AB \perp AC$ implies $m\angle BAC = 90^\circ$ - $DE \perp FG$ implies $m\angle GFE = 90^\circ$ - $CD \cong CE$ means triangle $CDE$ is isosceles with $CD = CE$ 3. **Find $m\angle C$ in triangle ABC:** Sum of angles in triangle ABC is $180^\circ$. $$m\angle A + m\angle B + m\angle C = 180^\circ$$ Since $m\angle A = 90^\circ$ and $m\angle B = 44^\circ$, $$90^\circ + 44^\circ + m\angle C = 180^\circ$$ $$m\angle C = 180^\circ - 134^\circ = 46^\circ$$ 4. **Use isosceles triangle property in triangle CDE:** Since $CD = CE$, angles opposite these sides are equal. 5. **Find $m\angle FGB$:** Given the perpendicular lines and congruent segments, $m\angle FGB$ corresponds to $m\angle C$ by alternate interior angles. Therefore, $$m\angle FGB = 46^\circ$$ **Final answer:** $$m\angle FGB = 46^\circ$$