# Three Isosceles Triangles Inside a Triangle +-- {.image} [[ThreeIsosceleseTrianglesInsideaTriangle.png:pic]] > The three shaded triangles are isosceles. What's the missing angle? =-- ## Solution by [[Angles at a Point]], [[Angles on a Straight Line]], [[Angles in a Triangle]], [[Angles in a Quadrilateral]] +-- {.image} [[ThreeIsoscelesTrianglesinaTriangleLabelled.png:pic]] =-- With the points labelled as in the diagram, let $a$ be angle $E \hat{O} A$, $b$ angle $B \hat{O} D$, and $c$ angle $A \hat{O} B$. Then as [[angles at a point]] add up to $360^\circ$, $c = 360^\circ - 35^\circ - a - b$. Triangle $E O A$ is [[isosceles]] so angle $A \hat{E} O$ is $\frac{180^\circ - a}{2} = 90^\circ - \frac{a}{2}$ as [[angles in a triangle]] add up to $180^\circ$. So since [[angles at a point on a straight line]] also add up to $180^\circ$, angle $O \hat{E} C$ is $90^\circ + \frac{a}{2}$. A similar argument shows that angle $C \hat{D} O$ is $90^\circ + \frac{b}{2}$. Since the [[angles in a quadrilateral]] add up to $360^\circ$, considering quadrilateral $O D C E$ gives the equation: $$ 50^\circ + 35^\circ + 90^\circ + \frac{a}{2} + 90^\circ \frac{b}{2} = 360^\circ $$ and this simplifies to $a + b = 190^\circ$. Hence $c = 135^\circ$.