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Geometry ukmt book question

Soo, I've been going trough the ukmt handbook "Plane Euclidean Geometry" by A.D. Gardiner and C. J Bradley and I am very stuck on one of the questions - Question 4 of 1g section. It basically asks to prove that if two medians are equal the triangle is isosceles. The problem is that I want to solve it using only the material from chapter 1 (parallel lines, triangle congruencess, Pythagoras theorem , areas of triangles). I can't think of a way to prove it without first proving that medians divide each other in 2:1 ratio, but in any proof of that I'm trying I need to use the midpoint theorem, which as far as I know is proved through similarities which are studied in chapter 2, so I would assume they are not intended to be used yet.. (I could do it with vectors, but probably it is also not the intended method) Anyone has suggestions that don't involve similarities?
Reply 1
Original post by Smile_more
Soo, I've been going trough the ukmt handbook "Plane Euclidean Geometry" by A.D. Gardiner and C. J Bradley and I am very stuck on one of the questions - Question 4 of 1g section. It basically asks to prove that if two medians are equal the triangle is isosceles. The problem is that I want to solve it using only the material from chapter 1 (parallel lines, triangle congruencess, Pythagoras theorem , areas of triangles). I can't think of a way to prove it without first proving that medians divide each other in 2:1 ratio, but in any proof of that I'm trying I need to use the midpoint theorem, which as far as I know is proved through similarities which are studied in chapter 2, so I would assume they are not intended to be used yet.. (I could do it with vectors, but probably it is also not the intended method) Anyone has suggestions that don't involve similarities?

It would help to see the question and the solution youre talking about. But one way would be to join the two midpoints defined by the medians, then you have a trapezium where the two diagonals are equal. So youd have two overlapping triangles with common base and equal length "diagonals". Dropping perpendiculars from the ends of top base to the bottom base to form right triangles means you can argue that the triangles are congruent using sas so isosceles trapezium and so isosceles triangle.

Tbh, the first trapezium part is the usual way to start prove the midpoint rule, as you probably know (then similar 1:2). So even if you dont want to use a result like the midpoint rule, its often worth learning/being inspired by the proof steps then thinking how they can be adapted for a similar problem.
(edited 2 months ago)
Reply 2
Original post by mqb2766
It would help to see the question and the solution youre talking about. But one way would be to join the two midpoints defined by the medians, then you have a trapezium where the two diagonals are equal. So youd have two overlapping triangles with common base and equal length "diagonals". Dropping perpendiculars from the end of top base to the bottom base to form right triangles means you can argue that the triangles are congruent using sas so isosceles trapezium and so isosceles triangle.
Tbh, the first trapezium part is the usual way to start prove the midpoint rule, as you probably know (then similar 1:2). So even if you dont want to use a result like the midpoint rule, its often worth learning/being inspired by the proof steps then thinking how they can be adapted for a similar problem.

Thank you so much for the reply! I'm a bit confused with the first part about the trapezium. How can I actually prove that the acquired quadrilateral is trapezium? Since I can't really use the fact that the segment between two midpoints is parallel to the base without proving it first. (what I initially referred to as midpoint theorem)
Reply 3
Original post by Smile_more
Thank you so much for the reply! I'm a bit confused with the first part about the trapezium. How can I actually prove that the acquired quadrilateral is trapezium? Since I can't really use the fact that the segment between two midpoints is parallel to the base without proving it first. (what I initially referred to as midpoint theorem)

Its a fairly trivial proof that its a trapezium (parallel top and bottom), or the original and upper (small) triangles are similar. It would be the same as the first part of the midpoint theorem proof, but the midpoint would use then go on to argue about the 1:2 ratio ... Here were just using the fact that the triangles are similar so the top/bottom is parallel so ...

So the vertex is common in the two triangles, then its just thales intercept theorem. So similar. Dont know the book but there must be something "similar" in that or a previous chapter? I guess you could simply state thales intercept theorem to get parallel without even mentioning the two triangles are similar / vertex is common, though its a fairly trivial observation.
(edited 2 months ago)
Reply 4
Original post by mqb2766
Its a fairly trivial proof that its a trapezium (parallel top and bottom), or the original and upper (small) triangles are similar. It would be the same as the first part of the midpoint theorem proof, but the midpoint would use then go on to argue about the 1:2 ratio ... Here were just using the fact that the triangles are similar so the top/bottom is parallel so ...
So the vertex is common in the two triangles, then its just thales intercept theorem. So similar. Dont know the book but there must be something "similar" in that or a previous chapter? I guess you could simply state thales intercept theorem to get parallel without even mentioning the two triangles are similar / vertex is common, though its a fairly trivial observation.

The contents page looks like this, and this problem is placed in 1.10, while similarity and proportion is only studied in 2.1, so I would assume similarities or any other theorems derived from them are not indented to be used yet(as this book basically teaches you geometry from zero)
WIN_20241209_12_45_30_Pro.jpg
Reply 5
Original post by Smile_more
The contents page looks like this, and this problem is placed in 1.10, while similarity and proportion is only studied in 2.1, so I would assume similarities or any other theorems derived from them are not indented to be used yet(as this book basically teaches you geometry from zero)WIN_20241209_12_45_30_Pro.jpg

Id guess the relevant stuff is in 1.7 but cant tell exactly whats in that section from the title. Id be a bit surprised if it didnt cover something like thales intercept. One way to prove it (claim 3) is using contradiction
https://en.wikipedia.org/wiki/Intercept_theorem#Claim_3
but it seems a bit much for the question. When I was talking about proportion in the previous post it was pretty much the defn of a median so it bisects the opp side so 1:1.

If you want to see how the parallel arguement would go for this book, id really need to see those few pages.
(edited 2 months ago)
Reply 6
Wow, thank you so much! Thanks to your help I have finally found + understood the solution for this question =)
Reply 7
Original post by Smile_more
Wow, thank you so much! Thanks to your help I have finally found + understood the solution for this question =)

Good. Out of curiosity, what argument did they make for the parallel lines?
Reply 8
Original post by mqb2766
Good. Out of curiosity, what argument did they make for the parallel lines?

The book just simply doesn't provide solutions to problems in chapters 1, 2 (although there are hints for some of them) so I can't say what the initially intended method was 😅. (Or do you mean the 1.7?)
Reply 9
Original post by Smile_more
The book just simply doesn't provide solutions to problems in chapters 1, 2 (although there are hints for some of them) so I can't say what the initially intended method was 😅. (Or do you mean the 1.7?)

Sure, I realise that. You seem to be now happy that the parallel lines stuff was not proven using stuff from chapter 1. I was just wondering why that was, so was there something like thales intercept theorem in 1.7 or ...
(edited 2 months ago)
Reply 10
Original post by mqb2766
You seemed to be happy that the parallel lines was now proved using stuff from chapter 1. I was just wondering why that was, so was there something like thales intercept theorem in 1.7 or ...

No, no intercept theorem. Quite literally the only things proved in 1.7 were (I simplified some of the wording):
1) if lines have any chance of being parallel, then two angles on one side of any transverse add up to exactly 180,
) two distinct lines are parallel if and only if transverse line creates interior angles on one side that add up to exactly 180,
3) if a point is not on the line m, there exists a unique line m' through that point that is parallel to m.

But the Claims 1 and 3 in the Wikipedia article that you provided, allowed to prove the midpoint theorem using only the knowledge given in chapter 1, with main ideas being to assume that two lines do intersect and then prove that its not the case - as was done in 1.7(although there are few similarities other than that) and areas from 1.8. I guess this book just expects a reader who is very quick to connect new concepts together, or maybe there is another solution where proving half of the intercept theorem just to solve the task is not necessary. (I hope I understood your question correctly haha)
Reply 11
Original post by Smile_more
No, no intercept theorem. Quite literally the only things proved in 1.7 were (I simplified some of the wording):
1) if lines have any chance of being parallel, then two angles on one side of any transverse add up to exactly 180,
) two distinct lines are parallel if and only if transverse line creates interior angles on one side that add up to exactly 180,
3) if a point is not on the line m, there exists a unique line m' through that point that is parallel to m.
But the Claims 1 and 3 in the Wikipedia article that you provided, allowed to prove the midpoint theorem using only the knowledge given in chapter 1, with main ideas being to assume that two lines do intersect and then prove that its not the case - as was done in 1.7(although there are few similarities other than that) and areas from 1.8. I guess this book just expects a reader who is very quick to connect new concepts together, or maybe there is another solution where proving half of the intercept theorem just to solve the task is not necessary. (I hope I understood your question correctly haha)

Id probably have to see the chapter to guess which way theyd want you to go. To get the side ratio, you could assume a parallel line passing through one side bisector point and then justify the side ratio for the other side using the sin rule, but that will likely be beyond chapter 1 and while sin rule is on the ukmt syllabus, its not "basic/chapter 1". Or thales intercept converse or .... (but theyd probably have covered forward thales). I dont think youre right about the book expecting you to quickly join together stuff, but ...?
Reply 12
Original post by Smile_more
No, no intercept theorem. Quite literally the only things proved in 1.7 were (I simplified some of the wording):
1) if lines have any chance of being parallel, then two angles on one side of any transverse add up to exactly 180,
) two distinct lines are parallel if and only if transverse line creates interior angles on one side that add up to exactly 180,
3) if a point is not on the line m, there exists a unique line m' through that point that is parallel to m.
But the Claims 1 and 3 in the Wikipedia article that you provided, allowed to prove the midpoint theorem using only the knowledge given in chapter 1, with main ideas being to assume that two lines do intersect and then prove that its not the case - as was done in 1.7(although there are few similarities other than that) and areas from 1.8. I guess this book just expects a reader who is very quick to connect new concepts together, or maybe there is another solution where proving half of the intercept theorem just to solve the task is not necessary. (I hope I understood your question correctly haha)

Actually, thinking about it a bit, you could do a construction with parallel lines and argue about lengths and angles and congruence. Ill sketch something tomorrow and upload it then.
Reply 13
Original post by Smile_more
No, no intercept theorem. Quite literally the only things proved in 1.7 were (I simplified some of the wording):
1) if lines have any chance of being parallel, then two angles on one side of any transverse add up to exactly 180,
) two distinct lines are parallel if and only if transverse line creates interior angles on one side that add up to exactly 180,
3) if a point is not on the line m, there exists a unique line m' through that point that is parallel to m.
But the Claims 1 and 3 in the Wikipedia article that you provided, allowed to prove the midpoint theorem using only the knowledge given in chapter 1, with main ideas being to assume that two lines do intersect and then prove that its not the case - as was done in 1.7(although there are few similarities other than that) and areas from 1.8. I guess this book just expects a reader who is very quick to connect new concepts together, or maybe there is another solution where proving half of the intercept theorem just to solve the task is not necessary. (I hope I understood your question correctly haha)

Id guess theyd have wanted you to do something like
Untitled presentation (4).jpg
So assuming D bisects AB (property of the median through <C) then use property 3 to construct two unique parallel lines DE and FE through D and E respectively. Then the aim is show E bisects AC. A simple chasing of angles and noting EF is the same length as DB (parallelogram), then its easy to show triangles ADE and EFC are congruent so E bisects AC ...

This is the converse so claim 1, but argueing about the uniqueness of E (like claim 3) its fairly trivial to flip the claim.
Reply 14
Original post by mqb2766
Id guess theyd have wanted you to do something like
Untitled presentation (4).jpg
So assuming D bisects AB (property of the median through <C) then use property 3 to construct two unique parallel lines DE and FE through D and E respectively. Then the aim is show E bisects AC. A simple chasing of angles and noting EF is the same length as DB (parallelogram), then its easy to show triangles ADE and EFC are congruent so E bisects AC ...
This is the converse so claim 1, but argueing about the uniqueness of E (like claim 3) its fairly trivial to flip the claim.

Thank you! This looks much more suitable for the task :> Although, I can't imagine how to even come up with this method, I guess it might come with practice
Reply 15
Original post by Smile_more
Thank you! This looks much more suitable for the task :> Although, I can't imagine how to even come up with this method, I guess it might come with practice

Id guess they wanted you to think about constructing parallel lines (so the third point with the uniqueness) and then arguing about congruence. Its a fairly standard ukmt thing to do/practice. There was a similar smc question a few years ago about a so triangle with 2 medians and asking about the area. Surprisingly the ukmt solution was to do the 2:1 midpoint calculation ... whereas it was a fair bit simpler to do the above construction and just multiply diagonals and note similarity for areas
https://www.thestudentroom.co.uk/showthread.php?t=7504510

As above though, there are several ways to prove it (as is often the case in geometry), depending on what you assume, so deriving stuff from euclids axioms. In retrospect, thales intercept theorem seems to be at about the right level for the problem, but it hasnt been covered yet in the book by the looks of it.
(edited 2 months ago)
Reply 16
Original post by mqb2766
Id guess they wanted you to think about constructing parallel lines (so the third point with the uniqueness) and then arguing about congruence. Its a fairly standard ukmt thing to do/practice. There was a similar smc question a few years ago about a so triangle with 2 medians and asking about the area. Surprisingly the ukmt solution was to do the 2:1 midpoint calculation ... whereas it was a fair bit simpler to do the above construction and just multiply diagonals and note similarity for areas
https://www.thestudentroom.co.uk/showthread.php?t=7504510
As above though, there are several ways to prove it (as is often the case in geometry), depending on what you assume, so deriving stuff from euclids axioms. In retrospect, thales intercept theorem seems to be at about the right level for the problem, but it hasnt been covered yet in the book by the looks of it.

I think the median ratio might be a bit more well known than the multiplication of diagonals trick for area so they chose it, although not sure. Thank you for the link, now I finally know some useful facts about area ^_^ (I think I've heard about them like 4+ times, but never was able to remember haha)
Reply 17
Original post by Smile_more
I think the median ratio might be a bit more well known than the multiplication of diagonals trick for area so they chose it, although not sure. Thank you for the link, now I finally know some useful facts about area ^_^ (I think I've heard about them like 4+ times, but never was able to remember haha)

The product of diagonals (halved and multiplied by sin(theta)) just falls out of the usual (ukmt) trick of splitting the quadrilateral into triangles using the diagonals, then factorising the area of each triangle. Its pretty much a one line proof (factorisation part). Its arguably simpler than the median-midpoint stuff, but easy to overlook. Like a fair bit of the ukmt stuff, its about being able to apply simple formulae to unusual problems.
Reply 18
Just having a flick through a few books and as well as the ukmt one
Four Pillars of Geometry - stillwell
is a quite readable companion to the ukmt stuff. It says its an undergraduate text (and maybe some parts are), but most of it is very readable. Slightly different order of presentation and Id probably argue more usual/better, but the ukmt one certainly seems to start at the basics.

A bit more of a "celebrations" book (so each short chapter is a 15 min read with a shiny pic)
Beautiful Geometry - maor
Ive an old "a level chapter" on the key greek stuff in a reasonably concise / readable form. Just message me if you want a copy.

There are plenty of others and theyre obv not ukmt focussed, but the previous two/three are pretty good for a keen person, but maybe see if you can get some form of preview (google books, sample chapter, pdf) to make sure you find them readable/accessible before handing over any hard earned cash. I find that often the commenty in a geometry book about why they do things is the most important thing. So
Excursions in geometry - ogilvy
is old and probably slightly above you, but very readable (and cheap). His excursions in number theory is a very good read and well worth the £1 I spent on it a few years ago.
(edited 2 months ago)

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