- 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:
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.
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.