Library / Berkeley Math Circle /18 of 98

Berkeley Math Circle

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.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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