Geometric Proof

Draw an equilateral triangle ABC. Bisect all three sides. Label the midpoints of AB, BC, and CA D, E, and F, respectively. Draw a line from D to E, and another from E to F. Draw a line from B to F and another from C to D. Draw a line from A to the intersection of BF and DE, and another from A to the intersection of CD and EF. These two lines intersect segment BC, dividing it into three peices. Show that these three segments are the same length, or find the ratio between them. What if the triangle is not equilateral?

(No, this isn’t homework.)

Sure sounds exactly like a homework problem to me. I haven’t even taken math a math course in years and I could figure this out if I knew it wasn’t going to be copied onto notebook paper later on. :stuck_out_tongue:

I haven’t taken an english course in a while either. :smack:

Didn’t Euclid say something about not trisecting an angle with compass and straightedge? Or was that a line? From my sloppy freehand sketch the segments look equal, but the ratio might be 1/4:1/2:1/4. Or something else.

Good luck with your homework!

(A suggestion anyway: Always look for symmetry, right triangles, or ratios you know already. Equilateral or right triangles are usually easiest to deal with.)

I haven’t taken a geometry course in three years. I’ve given this problem to a few people in my Calc III class and no one has figured out how to do it, so I’ve turned to the almighty Straight Dope for the answer. I don’t really know how to prove that it isn’t homework though.

I’ll take a stab at the initial problem, but it may comfort you to know that it doesn’t matter whether the triangle is equilateral or not. Ratios of the lengths of line segments are preserved by affine maps, and there is an affine map between any two triangles.

I should have realized that myself. My math homework last week was mostly about affine transformations and using them to simplify double integrals. :smack:

You answer lies in the use of each angle’s SIN, COS, and TAN. Or, just bust out that TI-85 and rip it up.

I’ve found that in a lot of these constructive geometric proofs, the trick is to “draw the missing line.” That is, there’s a line (or possibly more than one) which is not drawn in the original image but which is very useful for completing the proof. (Of course, there are a lot of possible lines to draw. Sometimes I end up drawing a lot of extra lines before I find the right one.)

In this case, one such line is the line joining the two intersection points. (I.e., let X be the intersection of DE and BF, and Y the intersection of CD and EF. Draw the line XY; let it intersect at M with AB and at N with AC.) This line is parallel to BC, so it completes a triangle AMN similar to ABC. What is the ratio of their side lengths? (And why should you care?)

Galois (effectively) said it’s impossible to trisect the angle.

Oh, if you don’t need it in Euclidean geometry, and you’re this far into a calc 3 class, this should be simple.

Here, I’ll hold your hand to start out: since it’s about proportions you can pick one side (say BC) and scale it down to length 1. Put B at (-1/2,0) and C at (1/2,0). Let A vary over the first quadrant and you’ll get a member of each similarity class of triangles.

From here it’s simple analytic geometry. RTFT.

I can outline the proof for you, but I’m too lazy to solve it all the way:

Let the point in the center of the equilateral triangle be X.
Let the intersection of FE and CD be Y.
Let |FA|=1.

Then the quadrilateral FAXY has sides
|FA|=1,
|AX|=(1/2)+1/(2sqrt3),
|XY|=1/(2sqrt3),
|YF|=1/2.
It has angles
YFA=120
FAX=30
AXY=120
XYF=90

You can get all that with various 30-60-90 triangles.

Now, using trig I’ve forgotten, you can get the diagonal AY since you have SAS on triangle YFA. You can crack it open, now, getting angle FYA. Let the intersection of FD and AY be Z. Now you can get angles YZF and FZA. Now you have ASA on triangle FAZ, and you can get the length of side FZ.

By similarity or whatever, FZ is half the length of one of the three pieces in the question. From that you can get the length of the other two pieces.

So basically, it is a geometry final in one problem.

Good luck.

So, let’s call the intersection of BF and DE, ‘G’. You probably know that bisects both segments.

Let’s call where AG intersect BC, ‘H.’

Let’s call where CD intersects EF, ‘I’.

Let’s call where AI intersects BC, ‘J’.

You want to know if BH=HJ=JC.

No.

Assume the length of the sides of the triangle are 1 unit =AB=BC=CA.

The midpoint of AC, which is F, creates the length, .5=AF.

So, for AFB, AB, the hypoteneuse = 1; AF = .5, therefore, BF = .866.

GF = .433 (half of BF)

AGF is a right triangle. The sine of GAF = .433/.5 = .866 = 40 degrees = HAC (also)

If BAF is 60 degrees and GAF is 40 degrees, this means BAH is 20 degrees.

If BAH is 20 degrees, by symmetry, JAC is 20 degrees.

If HAC is 40 degrees and JAC is 20 degrees, then HAJ is 20 degrees.

Congratulations, you’ve discovered how to trisect an angle (only works if the angle is 60 degrees).

Anyway, in triangle AJC, we know that JAC is 20 degrees and ACJ is 60 degrees, that means angle AJC is 100.

If we make the length of AC =1, then CJ should be .33…

However if we use our law of sines, the sine of AJC (100 degrees) divided by the length of AC (1 unit) should equal the sine of JAC (20 degrees) divided by CJ. Solving for CJ, we get ~.347

And so, CJ is not perfectly one third the length of BC, thus the three segments can not all be equal to each other.

It’s not sophisticated, but brute force is manageable here.

A = (0, 0)
B = (1/2, sqrt(3)/2)
C = (1, 0)
D = (1/4, sqrt(3)/4)
E = (3/4, sqrt(3)/4)
F = (1/2, 0)

G is the intersection of BF and DE, H is the intersection of CD and EF, I is the intersection of AG and BC, and J is the intersection of AH and BC.

G = (1/2, sqrt(3)/4)
H = (5/8, sqrt(3)/8)
I = (1/3, 2/sqrt(3))
J = (5/6, 1/(2sqrt(3)))

I get BI = sqrt(13)/6, IJ = (4 - sqrt(13))/6, and JC = 1/3. You can figure out the ratios yourself.

On the way home from work, I thought of an easier way to solve this. It’s really simple.

Going off Moriah’s diagram, extend GI to intersec DB and FC. Note that this is three segments of length .25. By similar triangles, that makes BH, HJ, and JC 1/3 each.

BTW, Moriah is confused at this step:

The sine of GAF = .433/.5 = .866 = 40 degrees = HAC (also).

That is nonsense. It’s opp/hyp, not opp/adj, and sin[sup]-1/sup=60, not 40.

Oops. You’re right, I used opp/adj, which is the tangent function. My calculations used tangent, but I typed ‘sine’ in my proof. Sorry.

The TANGENT of GAF = .433/.5 = .866 = 40 degrees = HAC (also).

Hmm… jawdirk’s solution is perfect. So why isn’t mine?

Recalculating… aha.

I rounded too much.

AGF is a right triangle. The TANGENT of GAF = .433/.5 = .866 = 40.9 degrees = HAC (also)

If BAF is 60 degrees and GAF is 40.9 degrees, this means BAH is 19.1 degrees.

If BAH is 19.1 degrees, by symmetry, JAC is 19.1 degrees.

If HAC is 40.9 degrees and JAC is 19.1 degrees, then HAJ is 21.8 degrees.

Congratulations, you’ve discovered how to trisect an angle (only works if the angle is 60 degrees). Ignore this.

Anyway, in triangle AJC, we know that JAC is 19.1 degrees and ACJ is 60 degrees, that means angle AJC is 100.9

If we make the length of AC =1, then CJ should be .33…

If we use our law of sines, the sine of AJC (100.9 degrees) divided by the length of AC (1 unit) should equal the sine of JAC (19.1 degrees) divided by CJ. Solving for CJ, we get ~.333…

And so, CJ is perfectly one third the length of BC, by symmetry, so is BH, and by subtraction, so is HJ.

So… yes.