Well I'm officially stuck. I'm reading through Birkhoff's

>>7769845
The area of the triangle is 1/2ck=1/2bh
QED

>>7769850
We haven't defined area yet, bub. Neither have we anything that shows that if two triangles have the same area they are similar.

File: swimming man.png (360KB, 457x362px) Image search: [Google]

360KB, 457x362px

>tfw grad student in pure math and dont know any of this kind of foundational geometry and it terrifies me

>>7769845
Get rid of AD in the diagram. It's useless and all it does is confuse you.

First, label the intersection point O.

Now compare some angles to see that ACF ~ CEO, and also that ABE ~ BFO

h/c = OF/OB
b/k = OC/OE
bh/ck = (OF*OC)/(OB*OE) = OF/OE*OC/OB

lastly now, EOC ~ BOF, so OF/OE = OB/OC and now your right side is one, i.e. bh=ck

Don't think there are any mistakes in that.

>>7769939
How is b OC for example? b is AC.

>>7769997

I haven't even been through the details of the other guy's thing, but it's clear that his argument involves comparing similar triangles, and for two ratios to be equal does not require that their components are equal, literally e.g. 3/6 = 1/2, etc.

I have a question for the guy though: how may we know that all three altitudes coincide at a single point O?

I smell Ceva's theorem

>>7769939
OP here.
Not quite. The triangles are similar as stated but h/c is not OF/OB, but FB/OB, since EB is similar to FB, not OF. AE is similar to OF.

So h/c = FB/OB.

Same with b/k, which is OC/EC, not OC/OE. OE is similar to AF.

so bh/ck = (FB*OC)/(OB*EC) = FB/OB * OC/EC.

Now since EOC ~ BOF this means that:
OB ~ CO
OF ~ EO
FB ~ EC

FB/OB * OB/FB = 1

which means bh = ck.

So your solution was correct, it just made a slight mistake with which line segments where similar to which.

This does however lead me to question just how you would come up with something like this. Now this isn't the first exercise in the book, but I guess it's the first non-trivial proof. I'm having a hard time just keeping all these similar triangles in mind, since they're not all similar to each other. But to help me with stuff like this in the future, since similar triangles is something I'm struggling to work with, even if I grasp the concept, is it proper to say that if AB ~ A'B', and CD ~ C' D', then AB/CB = A'B' / C'D'?

Any other advice for working with similar triangles, just practically?

>>7771854
Yes, similitudes preserve ratios (it is obvious for rigid motions, since they preserve distances and for homotheties, it is a consequence of Thales' theorem), as well as nonoriented angles.
Another thing to remember is that if two triangles ABC, A'B'C' have the same angle at some vertex, say <ABC = <A'B'C' and if the sides connected to that vertex are in the same proportion (AB/A'B' = AC/A'C'), then they are similar (I guess this follows from Al-Kashi)

>>7772005
That makes sense. Birkhoff and Beatley actually start out with the: "Case 1 of similarity: Two triangles are similar if an angle of one equals and angle of the other are the sides including these angles are proportional" as an axiom.

I was just surprised in the sudden spike in difficulty of these proofs compared to earlier ones, but I think it's just because I'm not used to dealing with ratios when dealing with similar triangles.

OP here again. Looking over this again there is an even simpler proof, that's more straight-forward. I'll keep practicing with similar triangles but it seems a key to solving these is finding a way to express different line segments in terms of each other.

bh / ck = (AC*BE)/(AB*CF) = AC/CF * BE/AB
Now we observe that ABE ~ ACF
With:
BE ~ CF, and
AB ~ AC.

So BE/AB = CF/AC and AC/CF * BE/AB becomes AC/CF * CF/AC, which is 1, proving that bh = ck.

>>7769845
triangles ABE and ACF are similar, since corresponding sides are proportionate b/k=c/h, and therefore bh=ck

It appears you've taken a few missteps. Hint: You need to get an integral, try finding the rate of gradient in <AB-<BC ;)