# Solution to the [[Subdivided Triangle III]] Puzzle +-- {.image} [[SubdividedTriangleIII.jpeg:pic]] > This isosceles triangle has been split into four smaller ones. What’s the missing angle? =-- ## Solution by [[Angles at a Point on a Straight Line]], [[Angles in a Triangle]], and Properties of [[Isosceles Triangles]] +-- {.image} [[SubdividedTriangleIIIAnnotated.jpeg:pic]] =-- 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, $$ 90^\circ - \frac{x}{2} = 4 x $$ meaning that $x = 20^\circ$.