A Geometric Proof of Brooks’s Trisection?

[UPDATE: we have a proof! I included it at the end of the blog post.]

Yesterday I was looking at a few methods of angle trisection.

For instance, I made this applet showing how to use the “cycloid of Ceva” to trisect an angle. (It is based on Archimedes’s neusis [marked straightedge] construction.)

I also found David Alan Brook’s College Mathematics Journal article “A new method of trisection.” He shows how you can use the squared-off end of a straightedge (or equivalently, a carpenter’s square) to trisect an angle. (This is a different carpenter’s square construction than the one I wrote about recently.)

To perform the trisection of $\angle ABC$ (see figure below), bisect the segment $BC$ at $D.$ Then draw the segment $DE$ perpendicular to $AB$ . Draw a circle with center $C$ and radius $CD.$ Next, arrange the carpenter’s square so that one edge goes through $B$ , one edge is tangent to the circle, and the vertex, $F$ , sits on $DE$ . Then $\angle ABF=\frac13\angle ABC$ .

brookstrisect

I made an applet to illustrate this trisection.

Brooks’s proof used trigonometry. My question is: Is there a geometric proof that $\angle ABF=\frac13\angle ABC?$ I spent a little while working on it yesterday and couldn’t find one. If you can, let me know!

UPDATE: We have a proof! Thank you Marius Buliga!

In the proof we are referring to the figure below. Let $\alpha=\angle ABF$ and $\beta=\angle CBF.$ Then it suffices to show that $\beta=2\alpha.$ Let $G$ be the point of tangency. Draw segment $CG$ and extend $FD$ to $J$ on $CG.$ Then draw segments $DG,$ $BJ,$ and $CF.$ Because $\angle CGF=90^\circ,$ lines $CG$ and $BF$ are parallel, and hence $BFCJ$ is a parallelogram. This implies that $D$ is the midpoint of the diagonal $FJ;$ so $DF=DJ.$ Moreover, Because $FJ$ is the hypotenuse of the right triangle $FJG$ and $DG$ is the median, $\angle DGF=\angle DFG.$ We see that $\angle DFG=\angle ABF$ so $\angle DGF=\alpha.$ Finally, because $DG$ is a chord of the circle, $\angle DCG=2\angle DGF;$ that is, $\beta=2\alpha.$

brookstrisectproof

Update 2: Andrew Stacey just sent me an another proof:

In the proof we are referring to the figure below. Let $\alpha=\angle ABF$ and $\beta=\angle CBF.$ Then it suffices to show that $\beta=2\alpha.$ Let $G$ be the point of tangency of the carpenter’s square and the circle. Because $\angle BEF=\angle BFG=90^\circ,$ $\angle DFG=\angle EBF=\alpha.$ Construct $CG.$ Because $FG$ is a tangent line and $CG$ is a radius of the circle, $\angle CGF=90^\circ.$ Because $\angle CGF=\angle BFG= 90^\circ,$ $BF$ and $CG$ are parallel. Thus $\angle BCG=\beta.$ Construct a point $H$ so that $BCGH$ is a parallelogram. Notice that $F,$ $B,$ and $H$ are collinear. Let $J$ be the midpoint of $GH$ . Construct the line segment $DJ$ and the circle with center $J$ and radius $GJ$ . Note that the circle passes through $G,$ $H,$ $D,$ and $F$ —the first three because $HJ=GJ=DJ,$ and the fourth because $\angle GFH=90^\circ$ and $GH$ is a diameter of the circle. Because $DJ$ is parallel to $FH,$ $\angle DJG=\beta.$ Finally, $\angle DJG$ is a central angle and $\angle DFG$ is an inscribed angle and both cut off the same arc of the circle; thus $\beta=\angle DJG=2\angle DFG=\alpha.$

brookstrisectproof2

9 Comments

pepe potamo says:

March 30, 2017 at 12:19 pm

I think there is no such proof, because there can’t be.
https://en.wikipedia.org/wiki/Angle_trisection

Dave Richeson says:

March 30, 2017 at 12:22 pm

It is impossible to trisect an arbitrary angle with a compass and straightedge, but if you allow certain other tools—like the carpenter’s square in this case—then angle trisection is possible. There is a proof that this works, but it uses trigonometry, not geometry.
puntomaupunto says:

March 30, 2017 at 1:19 pm

@pepe: as Dave says, this proof uses a tool different from straightedge and compass. In other words, without the carpenter’s square you cannot construct right angle BFG with the given constraints.

chorasimilarity says:

March 30, 2017 at 2:12 pm

BF || GC and the segments BD and DC have equal lengths. Now denote by J the intersection of the lines FD and GC. Then BFCJ is a parallelogram so the segments FD and DJ have equal length. The triangle FJG is right, therefore DG and FD have equal lengths. So the angles ABF, DFG and DGF are all equal. Use the triangle DCG to prove that the angle DCG is twice the angle DGF.

Dave Richeson says:

March 30, 2017 at 2:33 pm

Very nice! Thank you. I’ll update my post in a few minutes!
1. chorasimilarity says:
  
  March 30, 2017 at 3:52 pm
  
  Sorry for the double comment, i thought the computer ate my first one :) Very nice post and construction. From the proof I suspect the construction fails in interesting ways in hyperbolic or elliptic geometry.

chorasimilarity says:

March 30, 2017 at 2:19 pm

Denote by J the intersection of the lines FD and GC. The angles BFG and FGC are right, so BF is parallel with GC. Then BFCJ is a parallelogram, so FD and DJ have equal lengths. The triangle FJG is right so FD and DG have equal lengths, therefore the angles DFG and DGF are equal. Use that HGC is straight and the triangle DCG to prove that the angle DCG is twice the angle DGF.

ajakaja says:

March 30, 2017 at 3:29 pm

Hi, random amateur reader here.

I found it interesting to notice that it’s not the use of a carpenter’s square, specifically, that is needed in order to perform this construction — it’s the ability to find the point ‘F’, subject to the constraint that angle BFJ is a right angle. A carpenter’s square is one tool for doing that, but any method of ‘inverting’ the problem “BFJ is a right angle. solve for F” will do.

You can do this by binary-searching for the position of F on the line segment ED – but this amounts to constructing an infinite series of points and taking their limit, and that’s equivalent to allowing yourself to compute 1/3 of an angle by using an infinite series of bisections like 1/3 = 1/4+1/16+1/256…, which is a known way of trisecting angles anyway. Infinite series approximations are not allowed, of course, in normal compass+straightedge constructions, so this doesn’t clash with the impossibility proof.

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