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Geometry Difficulty 3.5 AMC 10/12 Find the answer Slovenia

The lines containing the chords $AC$ and $BD$ of a circle with radius $1$ intersect at an angle of $45^\circ$ . The length of the chord $AB$ equals $1$ (see figure).

What is the size of the angle $\angle CBD$ ?

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Solution

Let $S$ denote the centre of the circle. Then the triangle $SBA$ is equilateral since all of its sides have length $1$ . So, $\angle BSA = 60^\circ$ . By the Central Angle Theorem $\angle BCA = \frac{1}{2} \angle BSA = 30^\circ$ . The outer angle of the triangle is always equal to the sum of the non-adjacent inner angles, which implies $\angle CBD = 45^\circ + 30^\circ = 75^\circ$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.