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The problem:

c is a circle with a diameter AB. t is the tangent at the point B. Now C and D are two points on t and at different sides of B. I draw the line segments AC and AD, the point where AC crosses c is named E and the point where AD crosses c is named F. Show that the triangle ACF is similar to EDA.

I have made many approaches to this, but none of them is really helpful.

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    $\begingroup$ Are $C,D$ points on the tangent $t$? $\endgroup$ Commented Apr 28, 2013 at 11:18
  • $\begingroup$ @HagenvonEitzen yes, I forgot that $\endgroup$
    – D180
    Commented Apr 28, 2013 at 11:20

2 Answers 2

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Let $\alpha = \angle BAC$ and $\beta = \angle BAD$. Observe that $AC = AB / \cos \alpha$, $AD = AB / \cos \beta$ (follows from definition of $\cos$), $AE = AB \cdot \cos \alpha$ and $AF = AB \cdot \cos \beta$ (follows from graph of $r = \cos \theta$ in polar coordinates). Bearing in mind that $\angle A$ is common for both triangles, this is sufficient to show similarity.

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BAC would be similar to BAD because after cos, BAC would be inserted and so work out the rest of the equation and the answer would be: A= BAC+BAD.

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