Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.6 AIME Prove it United States

Problem:

Circles $k_{1}$ and $k_{2}$ intersect at points $A$ and $B$ . Line $l$ is the common tangent to these circles and it touches $k_{1}$ at $C$ and $k_{2}$ at $D$ such that $B$ belongs to the interior of the triangle $ACD$ . Prove that $\angle CAD + \angle CBD = 180^{\circ}$ .

Solution

Solution:

Here we use the fact that the angle between the tangent and the chord is equal to the peripheral angle corresponding to the chord (this fact is easy to verify and can be found in any book on geometry). We have $\angle CAB = \angle DCB$ and $\angle DAB = \angle DCB$ . Hence $\angle CAD = \angle CAB + \angle DAB = \angle DCB + \angle DCB = 180^{\circ} - \angle DBC$ . The statement now follows directly.

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