# In $\triangle ABC$ with $D$ on $\overline{AC}$, if $\angle CBD=2\angle ABD$ and the circumcenter lies on $\overline{BC}$, then $AD/DC\neq 1/2$

• 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.

• @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. – daedsidog Jan 4 at 15:15
• Whoops. ....... – 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.