# Argument with my brother about triangles. This is not homework; I'm 48.

September 8, 2017 7:02 PM Subscribe

I KNOW IT'S TRUE I JUST CAN'T PROVE IT.
I am having the world's dumbest argument with my brother about triangles. He says if you have two triangles, with two pair of proportional, corresponding sides, then the triangles

Bonus shame: I'm a frickin'

Here's what (I'm pretty sure) I know.

Pretend we have triangle ABC that corresponds with triangle DEF. So let's say you make angle A and angle D the same degrees, and side AB is some random proportion to side DE, and side BC is the same random proportion to side DF. Angle C could be acute and angle F could be obtuse, couldn't they? If you make the larger triangle's side long enough?

It works in my brain, but when I try to draw it, I fail, because I DON'T KEEP PROTRACTORS LYING AROUND MY HOUSE.

**have**to be similar. I say no, you*totally*can have two non-similar triangles with two proportional, corresponding sides.Bonus shame: I'm a frickin'

*tutor*, and my brother is a frickin'

*architect*. One of us is eventually going to fail one of our clients. The only thing keeping me sane is knowing that, if I'm wrong, it won't cause structural collapse.

Here's what (I'm pretty sure) I know.

Pretend we have triangle ABC that corresponds with triangle DEF. So let's say you make angle A and angle D the same degrees, and side AB is some random proportion to side DE, and side BC is the same random proportion to side DF. Angle C could be acute and angle F could be obtuse, couldn't they? If you make the larger triangle's side long enough?

It works in my brain, but when I try to draw it, I fail, because I DON'T KEEP PROTRACTORS LYING AROUND MY HOUSE.

Pretty sure your brother is right. If you go here and scroll down to Section 6.2 Similar Polygons it sounds like the exact thing you are arguing about.

posted by selfmedicating at 7:11 PM on September 8, 2017

posted by selfmedicating at 7:11 PM on September 8, 2017

I'm also in architecture, and I agree with your brother. Two angles, or one angle and two sides, and you've got similar triangles. Once you have one angle and two sides, the angles at the other two vertices of the triangle are pretty much locked in - that's kind of the basis of trigonometry, where you have ratios of sides relating to a right angle.

But since you don't want to believe architects, here's the wikipedia article on it. It sounds like you're interested in the third option there.

posted by LionIndex at 7:12 PM on September 8, 2017

But since you don't want to believe architects, here's the wikipedia article on it. It sounds like you're interested in the third option there.

posted by LionIndex at 7:12 PM on September 8, 2017

Here's a link specific to your question, with a helpful animation! http://www.mathopenref.com/similarsas.html

posted by Blue Jello Elf at 7:13 PM on September 8, 2017

posted by Blue Jello Elf at 7:13 PM on September 8, 2017

Using law of sines and law of cosines, you can determine the length of the third side from two adjacent sides and the angle between them. So I think once you specify that one angle is the same as its corresponding angle on the other triangle, and the two adjacent sides are proportional to the corresponding sides, then you've specified that the third side is also proportional.

posted by d. z. wang at 7:13 PM on September 8, 2017

posted by d. z. wang at 7:13 PM on September 8, 2017

You're correct - you're thinking of the angle-side-side non-identity. If the angle specified is

posted by zeptoweasel at 7:16 PM on September 8, 2017 [4 favorites]

**not**the one between the two corresponding sides, the triangles are not necessarily similar.posted by zeptoweasel at 7:16 PM on September 8, 2017 [4 favorites]

If you're just talking sides of the triangle and NOT angles, then you do need three sides to be similar. Two won't cut it.

posted by LionIndex at 7:17 PM on September 8, 2017

posted by LionIndex at 7:17 PM on September 8, 2017

Best answer: There is no ASS postulate. That's how I learned it.

posted by jessamyn at 7:17 PM on September 8, 2017 [15 favorites]

posted by jessamyn at 7:17 PM on September 8, 2017 [15 favorites]

Best answer: Instead of triangles with different sizes, start with ones with the same length measurments (ratio = 1).

-you can prove congruence with SAS, i.e. as you go around the triangle, a side, an angle that uses that side as one base, and the other side used by that angle.

-you can prove congruence with ASA, that is, a side with the two angles that use it as a base.

-you cannot prove congruence with SSA, that is, a side, the next side, and the angle opposite the original side.

You are correct.

posted by notsnot at 7:18 PM on September 8, 2017 [4 favorites]

-you can prove congruence with SAS, i.e. as you go around the triangle, a side, an angle that uses that side as one base, and the other side used by that angle.

-you can prove congruence with ASA, that is, a side with the two angles that use it as a base.

-you cannot prove congruence with SSA, that is, a side, the next side, and the angle opposite the original side.

You are correct.

posted by notsnot at 7:18 PM on September 8, 2017 [4 favorites]

Ah, right, the angle has to be between the two sides to work.

posted by LionIndex at 7:19 PM on September 8, 2017 [1 favorite]

posted by LionIndex at 7:19 PM on September 8, 2017 [1 favorite]

Except he's talking SSA which isn't a similarity theorem. You would need 2 angles, 3 sides or two sides with the included angle. It's almost like the law of sines ambiguous case...but for similarity.

posted by filthy light thief at 7:20 PM on September 8, 2017

posted by filthy light thief at 7:20 PM on September 8, 2017

Best answer:

He didn't say anything about an angle being the same. So,

posted by amtho at 7:47 PM on September 8, 2017 [6 favorites]

*He says if you have two triangles, with two pair of proportional, corresponding sides, then the triangles have to be similar. I say no...*He didn't say anything about an angle being the same. So,

**you're right**. _Just_ having two sides the the same proportions isn't enough. They have to have the same angle to each other. He left out the identical angle.posted by amtho at 7:47 PM on September 8, 2017 [6 favorites]

Response by poster: OKAY! I'M RIGHT! THAT'S AWESOME!

Now how do I draw my triangle ABC and my triangle DEF that prove it visually?

posted by tzikeh at 7:56 PM on September 8, 2017 [1 favorite]

Now how do I draw my triangle ABC and my triangle DEF that prove it visually?

posted by tzikeh at 7:56 PM on September 8, 2017 [1 favorite]

Response by poster: To the folks who said I'm wrong -- my brother is only talking about sides -- nothing about angles. So none of the SAS SSS AAA AAS stuff comes into play.

posted by tzikeh at 7:59 PM on September 8, 2017 [1 favorite]

posted by tzikeh at 7:59 PM on September 8, 2017 [1 favorite]

If you set the proportion=1, does that make it easier to draw?

posted by stobor at 8:09 PM on September 8, 2017 [1 favorite]

posted by stobor at 8:09 PM on September 8, 2017 [1 favorite]

If you only know that two sides are proportional, you don't have enough to say that they are similar triangles.

And for labeling, you want to say that AB is proportional to DE, and BC is proportional to EF, not DF. The order that the triangles are stated in should tell you which sides are corresponding.

As for drawing them, here's some illustrations of the Law of Sines - ambiguous case, or this single example. Now scale one triangle up or down, instead of making inset triangles with the same side measures.

(All credit to my tired math teacher wife for these insights, who hopes these are sentences.)

posted by filthy light thief at 8:09 PM on September 8, 2017 [1 favorite]

And for labeling, you want to say that AB is proportional to DE, and BC is proportional to EF, not DF. The order that the triangles are stated in should tell you which sides are corresponding.

As for drawing them, here's some illustrations of the Law of Sines - ambiguous case, or this single example. Now scale one triangle up or down, instead of making inset triangles with the same side measures.

(All credit to my tired math teacher wife for these insights, who hopes these are sentences.)

posted by filthy light thief at 8:09 PM on September 8, 2017 [1 favorite]

Response by poster:

Yes, that was a mis-type on my part. Thanks.

(I should have bet him money. All I have now are bragging rights about the dumbest argument ever conducted between middle-aged siblings on a Friday night.)

posted by tzikeh at 8:20 PM on September 8, 2017 [12 favorites]

*And for labeling, you want to say that AB is proportional to DE, and BC is proportional to EF, not DF. The order that the triangles are stated in should tell you which sides are corresponding.*Yes, that was a mis-type on my part. Thanks.

(I should have bet him money. All I have now are bragging rights about the dumbest argument ever conducted between middle-aged siblings on a Friday night.)

posted by tzikeh at 8:20 PM on September 8, 2017 [12 favorites]

If you had congruent angles between your two pairs of proportional sides, you could conclude similarity.

However, from your description, you do not. Look up the "ambiguous case" showing that ASS/SSA is not a general triangle congruence theorem. I believe this is what you were trying to describe in your original post. Scale one of those triangles by whatever ratio you like, and there's your drawing.

(However, depending on the specific angle and side lengths, you can sometimes make the congruence/similarity conclusion you want. It's possible to derive the condition using some trig, but I don't remember it off the top of my head. But again, SSA generally does not force congruence/similarity.)

posted by ktkt at 9:11 PM on September 8, 2017

However, from your description, you do not. Look up the "ambiguous case" showing that ASS/SSA is not a general triangle congruence theorem. I believe this is what you were trying to describe in your original post. Scale one of those triangles by whatever ratio you like, and there's your drawing.

(However, depending on the specific angle and side lengths, you can sometimes make the congruence/similarity conclusion you want. It's possible to derive the condition using some trig, but I don't remember it off the top of my head. But again, SSA generally does not force congruence/similarity.)

posted by ktkt at 9:11 PM on September 8, 2017

If it's any consolation, this dumb argument about triangles (not homework!) was the

posted by LuckySeven~ at 9:12 PM on September 8, 2017 [10 favorites]

**best**, funniest thing I've read in this entire sad, shitty week, so thanks for the laugh, tzikeh. (And I'm glad you won.)posted by LuckySeven~ at 9:12 PM on September 8, 2017 [10 favorites]

Okay, so somehow filthy light thief's post was invisible to me before I wrote mine, but clearly I concur with him.

posted by ktkt at 9:14 PM on September 8, 2017

posted by ktkt at 9:14 PM on September 8, 2017

My wife and I briefly talked through this and I agree with everyone else who said some variation of "if we're not talking about

posted by radiosilents at 9:21 PM on September 8, 2017 [1 favorite]

*angles*, then two triangles with proportional, corresponding*sides*don't have to be similar at**all**... excepting the fact that triangles in general are, by definition, similar".posted by radiosilents at 9:21 PM on September 8, 2017 [1 favorite]

Best answer:

Instead of working with two triangles whose sides are proportional with some arbitrary proportion, let's use the proportion of 1 (that is, same length).

Draw a long horizontal ray. From the start A of the ray, draw a line segment AB of some length (say 1) at some fixed angle up from horizontal (say, at 60 degrees up from horizontal.)

This segment is the first known side length of your triangle.

Now, you're going to imagine attaching a second line segment BC with a pivot to point B. Make your line segment long enough so that if it's hanging down from point B it crosses the horizontal ray (that is, C is below the ray by a chunk).

Now, imagine swinging segment BC back and forth. There will be two positions where C actually lies on the ray; one position, say at point C', where angle ABC' (at B) is less than a right angle and one position, say at C'', where the angle at B is more than a right angle.

Now, the two triangles ABC' and ABC'' obviously have two sides proportional---equal in this case---since the side AB never moved, and the side BC just swung back and forth. But the two triangles ABC' and ABC'' are very not congruent or similar, because they have different angles at B.

posted by leahwrenn at 9:57 PM on September 8, 2017 [2 favorites]

*Now how do I draw my triangle ABC and my triangle DEF that prove it visually?*Instead of working with two triangles whose sides are proportional with some arbitrary proportion, let's use the proportion of 1 (that is, same length).

Draw a long horizontal ray. From the start A of the ray, draw a line segment AB of some length (say 1) at some fixed angle up from horizontal (say, at 60 degrees up from horizontal.)

This segment is the first known side length of your triangle.

Now, you're going to imagine attaching a second line segment BC with a pivot to point B. Make your line segment long enough so that if it's hanging down from point B it crosses the horizontal ray (that is, C is below the ray by a chunk).

Now, imagine swinging segment BC back and forth. There will be two positions where C actually lies on the ray; one position, say at point C', where angle ABC' (at B) is less than a right angle and one position, say at C'', where the angle at B is more than a right angle.

Now, the two triangles ABC' and ABC'' obviously have two sides proportional---equal in this case---since the side AB never moved, and the side BC just swung back and forth. But the two triangles ABC' and ABC'' are very not congruent or similar, because they have different angles at B.

posted by leahwrenn at 9:57 PM on September 8, 2017 [2 favorites]

And on a re-read, if all you know is two sides, you don't know squat. Again, supposed the constant of proportionality is 1.

Take two sticks. Connect them by a pivot at one end. Now you have a V-shape where you can change the angle at the apex of the V.

Think about all the triangles you can make using that V as two of the sides! All of them have two pairs of proportional sides. Most of them are not congruent.

If you don't like the constant of proportionality equal to 1, then make yourself two proportional V shapes, and then vary the angles. Still mostly not similar, unless you happen to choose that same angle for both Vs.

(Am I misunderstanding the question?)

posted by leahwrenn at 10:03 PM on September 8, 2017 [5 favorites]

Take two sticks. Connect them by a pivot at one end. Now you have a V-shape where you can change the angle at the apex of the V.

Think about all the triangles you can make using that V as two of the sides! All of them have two pairs of proportional sides. Most of them are not congruent.

If you don't like the constant of proportionality equal to 1, then make yourself two proportional V shapes, and then vary the angles. Still mostly not similar, unless you happen to choose that same angle for both Vs.

(Am I misunderstanding the question?)

posted by leahwrenn at 10:03 PM on September 8, 2017 [5 favorites]

*Take two sticks. Connect them by a pivot at one end. Now you have a V-shape where you can change the angle at the apex of the V.*

Yes, someone should hand the architect a compass.

posted by Jon Mitchell at 10:23 PM on September 8, 2017 [6 favorites]

I think part of the confusion in this question is that there are two different questions! There's this:

and this

In the first question, we're saying that the ratio AB:DE is the same as the ratio BC:EF. In the second, we're saying that and ALSO that the angle at A agrees with the angle at vertex D.

The good news is, in both cases the triangle isn't determined up to similarity, so you're right! But the questions are really different in nature. In the first case, there are infinitely many possibilities for the shape of the triangle. In the second case, at most two.

posted by escabeche at 4:16 AM on September 9, 2017 [6 favorites]

*He says if you have two triangles, with two pair of proportional, corresponding sides, then the triangles have to be similar. I say no, you totally can have two non-similar triangles with two proportional, corresponding sides.*and this

*Pretend we have triangle ABC that corresponds with triangle DEF. So let's say you make angle A and angle D the same degrees, and side AB is some random proportion to side DE, and side BC is the same random proportion to side EF. Angle C could be acute and angle F could be obtuse, couldn't they? If you make the larger triangle's side long enough?*In the first question, we're saying that the ratio AB:DE is the same as the ratio BC:EF. In the second, we're saying that and ALSO that the angle at A agrees with the angle at vertex D.

The good news is, in both cases the triangle isn't determined up to similarity, so you're right! But the questions are really different in nature. In the first case, there are infinitely many possibilities for the shape of the triangle. In the second case, at most two.

posted by escabeche at 4:16 AM on September 9, 2017 [6 favorites]

I agree with escabeche. The source of the confusion is a squishy handle on the problem.

posted by SemiSalt at 5:42 AM on September 9, 2017 [1 favorite]

posted by SemiSalt at 5:42 AM on September 9, 2017 [1 favorite]

Best answer: An idea to draw 2 triangles that have proportional sides, use an equilateral triangle with side length of 1. For this triangle, the angle between any of the 2 sides is 60 degrees.

Now think of a right triangle with side lengths of 1 and the right angle between them. The hypotenuse of this triangle is:

1^2 + 1^2 = sqrt(2)

posted by bonofasitch at 1:45 PM on September 9, 2017

Now think of a right triangle with side lengths of 1 and the right angle between them. The hypotenuse of this triangle is:

1^2 + 1^2 = sqrt(2)

posted by bonofasitch at 1:45 PM on September 9, 2017

What your brother is claiming is that if you have two clocks with proportional hands, then the two times that they tell always have to be equal.

(i.e., ((clock 1 hour hand) : (clock 1 minute hand)) = ((clock 2 hour hand) : (clock 2 minute hand)))

... I'll have to ask my next architect this question, just to make sure ... ;)

(To be fair, I think this conundrum can only be down to a misunderstanding concerning the constraints)

posted by labberdasher at 12:41 AM on September 10, 2017

(i.e., ((clock 1 hour hand) : (clock 1 minute hand)) = ((clock 2 hour hand) : (clock 2 minute hand)))

... I'll have to ask my next architect this question, just to make sure ... ;)

(To be fair, I think this conundrum can only be down to a misunderstanding concerning the constraints)

posted by labberdasher at 12:41 AM on September 10, 2017

Response by poster: My brother has responded that I should have "known" that it was implied that the angles between the two congruent sides of each triangle were congruent. I told him he shouldn't design any three-dimensional structures for a while, and that he'll be getting

Thanks, everyone!

posted by tzikeh at 8:45 AM on September 10, 2017 [1 favorite]

*Geometry for Dummies*from me for his birthday.Thanks, everyone!

posted by tzikeh at 8:45 AM on September 10, 2017 [1 favorite]

This thread is closed to new comments.

posted by Brockles at 7:11 PM on September 8, 2017