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.

  • $\begingroup$ @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
  • 1
    $\begingroup$ Whoops. ....... $\endgroup$ – Blue Jan 4 at 15:17

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.