Line segment $\overline{AB}$ is perpendicular to line segment $OAC$, with $O$ and $C$ at different sides of the meeting point at $A$. The lengths of $\overline{AC} = \overline{AB}$ and $\overline{OA} = 2\overline{AB}$.

Line segment $\overline{CD}$ is also perpendicular to line segment $OAC$; the two line segments meet at $C$ and point $D$ is on the same side of line $OC$ as point $B$. The length of $\overline{CD} = 4\overline{AB}$.

(Sorry I have not provided a diagram; imagine on a 2-D grid a pair of right triangles one nested inside the other, with vertices at $(0,0),(3,0),(3,4)$ and $(0,0), (2,0), (2,1)$.)

Prove, using only classical plane geometry, that $\angle BOD = \angle AOB$.

It would be easy enough to show (using trigonometry) that the two angles are equal by calculating $$\tan (2 \angle AOB) = \frac{\frac12 + \frac12}{1- \frac12 \frac12} = \frac43 = \tan (\angle COD) $$ but that does not meet the requirement of "only classical plane geometry." One could of course prove the tangent of the sum of two angles formula using classical geometry, and then use that as a lemma in the proof of the desired statement, but I'm looking for a simpler and more direct proof.

Another cute proof using math outside of Euclid would be to consider point $O$ to be the origin of the complex plane, and take point $B = 2+i$. Then $$(2+i)^2 = 3+4i = C$$ and the result follows from de'Moivres theorem. Again, this is nice but it is not the pure geometry proof I am looking for.

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    $\begingroup$ Do you know the classical plane geometry result that if $D \in \overline{BC}$, then $\overline{AD}$ bisects the angle $CAB$ if and only if $\frac{BD}{AB} = \frac{CD}{AC}$? $\endgroup$ – Daniel Schepler Apr 4 at 18:33

Triangle $CDO$ is right with sides $3x,4x$ and $5x$, so inradius is $$ r={3x+4x-5x\over 2} = x$$

Since $B$ is at equal distance from $OC$ and $CD$ we see that $B$ is on angle bisector for $\angle OCD$ and since this distance is $x=r$ we see that $B$ is an incenter and thus a conclusion.

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    $\begingroup$ Interesting, I get a down vote! Why? $\endgroup$ – Aqua Apr 4 at 19:17
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    $\begingroup$ Now this is depresive. For a completly correct solution, you even get a downvote. $\endgroup$ – Aqua Apr 4 at 19:40
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    $\begingroup$ The use of the incenter is the clever idea I was looking for. $\endgroup$ – Mark Fischler Apr 4 at 19:44

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