Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it South Africa

Let $A$ , $B$ , and $C$ be distinct points on a straight line with $AB = AC = 1$ .
Square $ABDE$ and equilateral triangle $ACF$ are drawn on the same side of line $BC$ .
Lines $EC$ and $BF$ cut in $G$ .
What is the size of $\angle EGB$ ?

Solution

We will compute the angle sizes of quadrilateral $DEGB$ and use the fact that the sum of the interior angles of a quadrilateral is equal to $360^\circ$ .

Since $AFC$ is an equilateral triangle, $\angle FAC = 60^\circ$ and so $\angle EAF = 90^\circ - 60^\circ = 30^\circ$ , which yields $\angle BAE = 90^\circ + 30^\circ = 120^\circ$ . Since $AB = AC = AF$ , triangle $BAF$ is isosceles, and so
$\angle GBC = \angle FBA = \frac{1}{2}(180^\circ - \angle BAF) = \frac{1}{2}(180^\circ - 120^\circ) = 30^\circ.$
Next, $AE = AB = AC$ , and so triangle $EAC$ is a right-angled isosceles triangle, which means that $\angle BCG = \angle ACE = 45^\circ$ . Hence,

$\angle$ BGE = $\angle$ GBC + $\angle$ GCB = $30^\circ + 45^\circ = 75^\circ.$

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