Proportional Segments Theorem

[noowutah]

I remember learning this in high school, but I can't track down a proof. Let ABC be a triangle and DE a line segment intersecting the triangle such that D is on AB, E is on AC, and DE is parallel to BC. Then

\frac{\overline{DE}}{\overline{BC}}=\frac{\overline{AD}}{\overline{AB}}=\frac{\overline{AE}}{\overline{AC}}

I duckduckgo'd this as much as I could, but no luck. There is a claim for a proof here

http://ceemrr.com/Geometry1/ParallelSimilar/ParallelSimilar_print.html

and I get the proof for the Triangle Midsegment Theorem, but I don't know what they mean when they say that the Proportional Segments Theorem follows from the Triangle Midsegment Theorem by "repeated application." How so?

 

[.Scott]

You can demonstrate that the two triangles are similar based on AAA (angle, angle, angle) - they share one angle and the parallel side can be used to prove the other two angle are equivalent.

The lengths of corresponding sides of similar triangles follow the same proportions.

 

[noowutah]

True. But say I can't use trigonometry (Pythagoras is OK) -- how do you show that the sides of congruent triangles are proportional? In my case below, it would be ideal to show that the rectangle with the sides DE and AB equals in area the rectangle with the sides BC and AD. For the theorem of Pythagoras, there is an elegant proof that the square c^2 equals the sum of the two squares a^2 and b^2. Similarly, for the Triangle Midsegment Theorem (see the link above). I don't see anything like that for the Proportional Segments Theorem, except an appeal to intuition.

 

[.Scott]

But say I can't use trigonometry (Pythagoras is OK) --

O...K... ?!?

how do you show that the sides of congruent triangles are proportional?

Am I allowed to used analytical geometry? That would give me the "Law of Sines".

In my case below, it would be ideal to show that the rectangle with the sides DE and AB equals in area the rectangle with the sides BC and AD.

Once you use the Law of Sines to prove the lengths of the corresponding sides are proportional, this would be a simple next step or two.

 

[Mark44]

You can demonstrate that the two triangles are similar based on AAA (angle, angle, angle) - they share one angle and the parallel side can be used to prove the other two angle are equivalent.

The lengths of corresponding sides of similar triangles follow the same proportions.

True. But say I can't use trigonometry (Pythagoras is OK) -- how do you show that the sides of congruent triangles are proportional? In my case below, it would be ideal to show that the rectangle with the sides DE and AB equals in area the rectangle with the sides BC and AD. For the theorem of Pythagoras, there is an elegant proof that the square c^2 equals the sum of the two squares a^2 and b^2. Similarly, for the Triangle Midsegment Theorem (see the link above). I don't see anything like that for the Proportional Segments Theorem, except an appeal to intuition.

What .Scott is suggesting is not trigonometry - it's plain old geometry. Since DE || BC, ∠ABC = ∠ADE (corresponding angles of parallel lines cut by a transversal). Exactly the same argument can be made to show that ∠AED = ∠ACB. This shows that the three angles of the smaller triangle are congruent to the corresponding angles of the larger triangle - this the triangles are similar.

Regarding this question -- "how do you show that the sides of congruent triangles are proportional" The corresponding sides of congruent triangles are equal in length, so technically they are proportional, with the proportion being 1. Perhaps you meant "sides of similar triangles" instead.

 

[noowutah]

Yes, pardon me, I meant similar triangles, not congruent triangles. And no, I don't want to refer to the law of sines for a proof. My question is if I can show by a plain geometrical argument that makes sense to a 12-year old (no trigonometry) that the sides of similar (!) triangles are proportional. There are such plain geometrical arguments for the Theorem of Pythagoras and for the Triangle Midsegment Theorem. I as yet can't think of one for the Proportional Segments Theorem without recourse to trigonometry.

Mentor's argument only shows me that the two triangles are similar, but not that their sides are proportional. Scott's argument uses trigonometry (the law of sines). By "OK" I mean it's OK to use P.'s theorem in your proof, but please no trig.

 

[SteamKing]

Yes, pardon me, I meant similar triangles, not congruent triangles. And no, I don't want to refer to the law of sines for a proof. My question is if I can show by a plain geometrical argument that makes sense to a 12-year old (no trigonometry) that the sides of similar (!) triangles are proportional.

Would this 12-year old be stuck doing some homework?

Try this link:

http://www.mathopenref.com/similartriangles.html

If you google 'similar triangles', you can find other similar (no pun intended) links.

 

[noowutah]

No, it's not homework. If it were, it appears that not only 12-year-olds, but also some physicists on physicsforums.com would be stumped, including me :)

The website you are linking to (I found it earlier as well) does a great job appealing to our intuition, but the proof is still missing. Again, one thing that would satisfy me is if you could show me that the rectangle with the sides DE and AB equals in area the rectangle with the sides BC and AD.

 

[Mark44]

It's been a long, long time since I took geometry in 10th grade, but I believe there was a theorem that said that similar triangles have sides that are proportional in the same ratio. Being a geometry book, it would not have used trigonometry to reach this result. Off the top of my head I don't remember how it was done.

 

[noowutah]

I am beginning to think that this isn't so easy ...

 

[.Scott]

Here you go.
The meat starts at the bottom of page two, but read it from the start.
If you're explaining it to a 12-year old, strike the parts about non-Euclidean geometry.

http://www.math.washington.edu/~lee/Courses/444-5-2008/supplement3.pdf

 

[noowutah]

Brilliant. That's it. And not so easy ...