Question: User: H A B I G F C D E Note: Figure not drawn to scale. In the figure shown, triangle ABC is an isosceles triangle, the length of CF is equal to the length of CD, and the length of FE is equal to the length of DE. If the measure of angle HAC is 100° and the measure of angle GFC is 40°, what is the measure of angle E, in degrees?

1. **State the problem:** We are given an isosceles triangle ABC with $CF = CD$ and $FE = DE$. We know $\angle HAC = 100^\circ$ and $\angle GFC = 40^\circ$. We need to find $\angle E$. 2. **Analyze the figure and given information:** - Since $ABC$ is isosceles, $AB = BC$ or $AC = BC$ (usually the equal sides are specified, but here we assume $AB = BC$ since $C$ is the vertex where $CF$ and $CD$ meet). - $CF = CD$ means triangle $CFD$ is isosceles with $CF = CD$. - $FE = DE$ means triangle $FED$ is isosceles with $FE = DE$. - $\angle HAC = 100^\circ$ is an angle on the horizontal line $H A B I$. - $\angle GFC = 40^\circ$ is an angle at point $F$. 3. **Use the given angles to find other angles:** - Since $\angle HAC = 100^\circ$ and $H A B I$ is a straight line, $\angle CAB = 80^\circ$ because $\angle HAC + \angle CAB = 180^\circ$. 4. **Find angles in triangle ABC:** - Triangle $ABC$ is isosceles with $AB = BC$. - We have $\angle CAB = 80^\circ$, so $\angle ABC = 80^\circ$. - The third angle $\angle BCA = 180^\circ - 80^\circ - 80^\circ = 20^\circ$. 5. **Analyze triangle CFD:** - $CF = CD$ so $\triangle CFD$ is isosceles with $\angle CFD = \angle CDF$. - $\angle GFC = 40^\circ$ is given. - Since $G, F, C$ form an angle of $40^\circ$ at $F$, and $CF = CD$, the angles at $F$ and $D$ in triangle $CFD$ are equal. - Let $\angle CFD = \angle CDF = x$. - The sum of angles in $\triangle CFD$ is $180^\circ$, so $x + x + 40^\circ = 180^\circ$. - Thus, $2x = 140^\circ$ and $x = 70^\circ$. 6. **Analyze triangle FED:** - $FE = DE$ so $\triangle FED$ is isosceles with $\angle FED = \angle EFD$. - We want to find $\angle E = \angle FED$. - Note that $\angle EFD$ is adjacent to $\angle CFD = 70^\circ$. - Since $F$ lies on the line segments $CF$ and $FE$, the angles around $F$ sum to $360^\circ$. - The angles at $F$ are $\angle GFC = 40^\circ$, $\angle CFD = 70^\circ$, and $\angle EFD$. - So, $\angle EFD = 360^\circ - 40^\circ - 70^\circ -$ (angle between $FE$ and $FG$ which is zero here) $= 250^\circ$ is impossible. 7. **Reconsider the angle sum at $F$:** - Actually, $\angle EFD$ and $\angle CFD$ are adjacent angles on a straight line $C-F-E$, so $\angle EFD + \angle CFD = 180^\circ$. - Since $\angle CFD = 70^\circ$, then $\angle EFD = 110^\circ$. 8. **Find $\angle E$ in triangle FED:** - Since $FE = DE$, $\triangle FED$ is isosceles with base $FD$. - Let $\angle FED = \angle EFD = y$. - Sum of angles in $\triangle FED$ is $180^\circ$, so $y + y + 110^\circ = 180^\circ$. - Thus, $2y = 70^\circ$ and $y = 35^\circ$. **Final answer:** $$\boxed{35^\circ}$$