Notes
subdivided triangle iii solution

Solution to the Subdivided Triangle III Puzzle

Solution by Angles at a Point on a Straight Line, Angles in a Triangle, and Properties of Isosceles Triangles

Consider the points labelled as above. Let $x$ be the angle at the apex, namely $C \hat{A} D$. Then as triangle $A D C$ is isosceles, angles $A \hat{D} E$ and $D \hat{C} A$ are both $90^\circ - \frac{x}{2}$.

Since angles in a triangle add up to $180^\circ$, and triangle $A F B$ is isosceles, angle $A \hat{F} B$ is $180^\circ - 2 x$. Then as angles at a point on a straight line also add up to $180^\circ$, angle $B \hat{F} E$ is $2 x$.

Then angle $E \hat{B} F$ is $180^\circ - 4 x$, so as angle $F \hat{B} A$ is $x$, this leaves angle $E \hat{B} C$ as $3 x$. Therefore also angle $E \hat{C} B$ is $3 x$.

Triangle $E C D$ is isosceles, and shares angle $E \hat{D} C$ with triangle $D A C$, so these triangles are similar, meaning that angle $E \hat{C} D$ is the same as angle $C \hat{A} D$, namely $x$. So angle $D \hat{C} A$ is $x + 3 x = 4 x$.

Therefore,

meaning that $x = 20^\circ$.