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enter image description here

  • Let $\alpha$ be $\measuredangle ABD$
  • Let $\beta$ be $\measuredangle DBC$
  • Let D be a point on AC such that BD passes through the origin point O

Prove that $\frac{AD}{DC}$ cannot be equal to $\frac{1}{2}$ when $\alpha = \frac{1}{2}\beta$

Here's what I have:

$$\measuredangle AOD = 2\alpha$$ $$\measuredangle DOC = 2\beta$$

From this information I completed all the angles and reached the following (correct) equation by using the law of sines on triangles AOD and DOC, which share the equal length OC = AO:

$$\frac{AD}{DC} = \frac{\sin 2\beta}{\sin 2\alpha}$$

Given that $\alpha = \frac{1}{2}\beta$:

$$\frac{AD}{DC} = \frac{\sin 2\beta}{\sin \beta} = \frac{2\sin \beta \cos \beta}{\sin \beta} = 2\cos\beta$$ $$\downarrow$$ $$2\cos\beta = 1/2$$ $$\downarrow$$ $$\beta = 75.52^\circ$$

If $\beta = 75.52^\circ$, then $2\alpha + 2\beta > 180^\circ$, and the triangle will now look like this:

enter image description here

In this situation, BD cannot pass through O, which breaks the definition of the problem.

This is the best proof I can come up with. I tried first to prove it numerically by summing up angles to 180, but that did not work as all the statements were true. I feel that my proof is borderline illegal and that I do not address all cases, so I am asking if anyone could figure out a more elegant, preferably algebraic alternative.

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  • $\begingroup$ @Blue If AC is the diameter, how is $AD/DC = 1/2$? Both AD and DC become radii of the circle and therefore their ratio should be 1. $\endgroup$ – daedsidog Jan 4 at 15:15
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    $\begingroup$ Whoops. ....... $\endgroup$ – Blue Jan 4 at 15:17
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If $\frac{AD}{DC}=\frac{1}{2}$ by the law of sines we obtain: $$\frac{1}{2}=\frac{AD}{DC}=\frac{\frac{AD}{BD}}{\frac{DC}{BD}}=\frac{\frac{\sin\alpha}{\sin\measuredangle A}}{\frac{\sin2\alpha}{\sin\measuredangle C}}=\frac{\frac{\sin\alpha}{\sin(90^{\circ}-2a)}}{\frac{\sin2\alpha}{\sin(90^{\circ}-\alpha)}}=\frac{\sin\alpha\cos\alpha}{\cos2\alpha\sin2\alpha}=\frac{1}{2\cos2\alpha}.$$ Id est, $\cos2\alpha=1,$ which is impossible.

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  • $\begingroup$ Strictly out of interest, could you share the thought process which lead you to this? $\endgroup$ – daedsidog Jan 4 at 15:22
  • $\begingroup$ @daedsidog It's law of sines. Which step is not clear? $\endgroup$ – Michael Rozenberg Jan 4 at 15:25
  • $\begingroup$ It's perfectly clear, I was asking for clues on how would one even realize to use law of sines in this case. $\endgroup$ – daedsidog Jan 4 at 15:26
  • $\begingroup$ @daedsidog See the second step (the second equlity). It's a preparation to using of the law of sines. $\endgroup$ – Michael Rozenberg Jan 4 at 15:30

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