I noticed this old post and came up with a different proof. The challenge was: With triangle ABC extend the base line AB in either direction. Let the angle bisector at A intersect the EXTERNAL angle bisector at B at point E, and similarly let the angle bisector at B intersect the external angle bisector at A at point D. Show that D, C, and E are collinear.
My proof goes like this: Note that DAE and DBE are right triangles. Therefore the circle circumscribing DAE also contains B. The arc EB is twice the inscribed angle, so it equals A, and the arc DA is B, so the arc AB is pi-A-B, which is the same as the angle at C. Therefore C must be on the diameter DE.
Relevant column: Can Cecil solve this geometry stumper? - The Straight Dope
The GIF with the illustration on the old column is missing, without which neither the original column nor this thread can be followed.
You underestimate Dopers. It’s pretty easy to construct with the instructions given. Although it is a bit tricky since Cecil didn’t clearly define how to place W, step 9 makes it clear, even if you need extra points and line segments.
Here’s a quick(ish) version I made.
And from what I remember of geometry, the OP needs to give us a reason why DAE and DBE are right triangles. (If I could remember the theorems, I’d help.)
I do believe that he is right about those two angles being right angles. The external and internal angles of each add up to 180 degrees, thus half of each angle would add up to 90 degrees. Like I said, I don’t remember the theorem names.
What I don’t know is if that means you can create a circle DABE. It seems you could, but I don’t remember any of the theorems about circles, even without names.
Well, you hardly need a name; you’ve given the reasoning! It’s so simple and direct that referring to it instead by some arcane name would only be more obfuscatory than helpful.
Almost everything the OP says is in fact true, except…
The very last sentence. It doesn’t make sense to me. The OP correctly demonstrates that DABE lies on a circle with diameter DE, and that the angle of the arc AB on that circle is the same as the angle ACB. But I don’t know how they conclude from this that C lies on on the diameter of that circle. I could easily construct counterexamples to the principle “If the angle ACB is the same as the angle of arc AB along some circle, then C lies on the diameter of that circle”. So I have no idea what is being invoked here. Could the OP (or someone) clarify?
In case anyone is interested in the original diagrams, I have (temporarily) posted a scan of the column, as it appeared in Return of the Straight Dope, at: http://www.ipass.net/whitetho/column.gif .
I realize I made some mistakes in my original approach to the geometry problem. Instead I have come up with an algebraic (analytic geometry ) proof of the theorem.
First, let the vertex A at the base be at the origin (0,0). Let the other base vertex be at (b,0). Now any triangle with base AB is formed by a line through (0,0) making an angle with the x-axis and another line through (b,0) making a second angle with the x-axis (which will actually be the “exterior” angle at B). Suppose we look at the line y=mx through (0,0) with slope m and the line y= n(x-b) through (0,b) with slope n. These intersect at the point (nb/(n-m),mnb/(n-m)). If we double the angles of the two lines, the new slopes are 2m/(1-m2) and 2n/(1-n2), by the double angle formula for tangents. Let’s assume these “double angle” lines are the ones which intersect at the original vertex C. If we rotate the two non-double angle lines by 90 degrees counterclockwise, the new slopes are -1/m and -1/n.
From the above, lines with slopes M and N intersect at (Nb/(N-M),MNb/(N-M)) and two other lines with slopes K and L intersect at (Lb/(L-K),KLb/(L-K)). A straightforward but lengthy calculation shows that the slope of the line between these two points is (KL(N-M)-MN(L-K))/(L(N-M)-N(L-K)). If the first pair of lines is the one with slopes M=m and N=n, and the second the double-angle lines with slopes K=2m/(1-m2) and L=2n/(1-n2), then line between them, which goes through the vertex C, by the above formula, has slope (mn-1)/(m+n). Similarly, if we use the lines with slopes M=m and N=n and the perpendicular lines with slopes K=-1/m and L=-1/n, then the above formula gives the same value, (mn-1)/(m+n), for the slope of the line through the two points where each pair of lines intersect. Therefore these lines are collinear.
Recall again that we are relating the vertex C where the doubled angle lines intersect with the vertex where the non-doubled lines intersect, and also with the vertex where the 90 degree rotated lines intersect. 90 degrees comes from the fact that adjoining a bisected angle with the bisected supplementary (exterior) angle gives half of a straight angle, or 90 degrees.
Please also see my next post.
From Cecil’s column:
I also was trying to take a holistic view of this problem. Thinking of the (non-base) sides of the triangle as lines through A and B making angles alpha and beta with the x-axis (beta actually being the “exterior angle”, then if we halve the angles to get a new vertex C’, the slope of the line through C and C’ is a function of alpha and beta, and, in some vague sense, of the two original lines themselves. Now if instead we rotate lines AC and BC by 180 degrees (giving actually the same lines) and halve these angles, we get the angle bisector at B and the exterior angle bisector at A, intersecting at C’’. So in some sense this set is also the original lines AC and BC with “their angles halved.” Therefore we might expect that the slope from C to C’’ is the same as the slope from C to C’. Perhaps this reasoning can be made more precise.
Simply talking about the triangle ADE is not problematic; it is, after all, some triangle. It would only be problematic if the OP had somewhere assumed that C lay on the edge between D and E, but I don’t see that they have mistakenly made that assumption anywhere (though the OP’s last step is still unjustified).
Actually, I think this is a geometric proof along the lines of my first comment:
Let AB be the base of the triangle, C the upper vertex. Let the angle bisector at A intersect the external angle bisector at B at point E, and let the angle bisector at B intersect the external angle bisector at A at point D. DEB and DEA are both right triangles with the same base DE, so a circle can be drawn through D, A, B, and E with diameter DE. Let the line BC intersect this circle at F, and the line AC intersect this circle at G. Then the arcs AD and DF both equal the angle B, the arcs BE and EG both equal the angle A, and the arcs AB and FG therefore must both equal the angle C. FGC and ABC are similar triangles, actually congruent since the chords FG and AB are equal. Therefore C is equidistant from G and B and so must be on the diameter DE, i. e. it is collinear with D and E.
Can you explain how knowing the two triangles are right triangles allows us to know that all four points circumscribe a circle? It makes sense intuitively, but I can’t seem to come up with how it works–not that I’ve tried especially hard.
The basic principle (which I will prove in a second) is that, for any right triangle, the circle whose diameter is the hypotenuse also contains the point opposite the hypotenuse. Thus, once we know that DAE and DBE are right triangles, we know that the circle whose diameter is DE also contains A and also contains B.
Ok, now why is this basic principle true? Well, pick any line segment you like, and for convenience, let us choose its midpoint to be our origin (thus, we can think of the two endpoints of the segment as -V and +V).
What is the equation describing the circle for which this line segment is the diameter? Well, P lies on that circle just in case P * P = V * V, where * is dot product.
What is the equation describing the points which can serve as the opposite corner of a right triangle with this line segment as hypotenuse? Well, P can do that just in case (P + V) * (P - V) = 0, where * is dot product.
But these two conditions are equivalent (the dot product acts just like ordinary multiplication, in that it is symmetric and distributes over addition/subtraction), and so a point may serve as the opposite corner of a right triangle with a particular hypotenuse just in case it lies on the circle whose diameter is that hypotenuse. Q.E.D.
I was using throughout my discussion the theorem that an angle inscribed in a circle is equal in measure to half the intercepted arc. (Remember that from geometry class?) Both right angles intercept the upper 180 arc and so the vertices A and B are on the lower semicircle. This also shows why the arcs opposite the various equal half angles are equal.