Theorem in Euclidian geometry

berliner · 2013-04-02 01:01:27

Hi everyone! I want to submit you a theorem in geometry.

Given a line and a point not on the line and two oblique carried out from the same point, distances from the midpoints of the oblique to the line are congruent.
Hypothesis: P doesnt belong to r wile H, A, R belong to r and PH is perpendicular to r. M is the midpoint of PA and N is the midpoint of PR.
Thesis: MS is congruent to NT
Begin to build the axis of symmetry t passing through the midpoint of MS.
St: S<->M.
r <-> r' and M belongs to r'
St: A <-> x. x is a point which belongs to r' because T and x correspond in an isometry which by definition maintains unchanged the distance of the points.
We must demonstrate that x is the midpoint of PR.
The symmetry transforms PR in P'R'. The midpoint of P'R' is T, because St: R<->R', P <-> P', PR<->P'R', PR = P'R' and NT is perpendicular to r at point T. So, since that T is the midpoint of P'R' and N is the midpoint of PR and since that PR and P'R' correspond then T and N correspond. So x = N and since x belongs to r' then also N belongs to r'. So M and N belong to r', S and A belong to r, r and r' are parallel (because parallelism is am invariant of symmetry) and MS and NT are perpendicular to r the MS and NT are congruent (because the distance between parallel lines doesn't change).

Bob · 2013-04-02 05:50:11

hi berliner,

Welcome to the forum.

Nice proof, but if I've understood you correctly, Euclid got there first. ref Elements book 6, converse of proposition 2.

http://farside.ph.utexas.edu/euclid/Elements.pdf

Bob

berliner · 2013-04-02 06:06:50

Thanks Bob;
one of the unpleasant things of mathematics is that sometimes someone is a step ahead of you.
I will try to demonstrate something else. At least mine is an alternative demonstration

Bob · 2013-04-02 06:15:18

hi berliner,

Is it unpleasant? I think it happens a lot. Newton and Leibniz spent a lot of their lives arguing about who invented calculus.

Personally, I'm just happy to understand some maths, and if I independently come up with something, to me that's still a discovery.

Good luck with your proving ....

Bob

berliner · 2013-04-02 06:47:10

You're right, Bob. Thanks!

berliner · 2013-04-03 03:00:14

Another theorem! I hope this one has not been demonstrated yet

Hypothesis: P, S ∉ r ∧ PH ⊥ r
H, R ∈ r ∧ HS ∩ PR = {M}
Thesis: ∠SMR > ∠MRH

Let's indicate with π the straight angle. ∠SMR +∠HMR = π . ∠HMR belongs to the triangle MRH. So, since the sum of the interior angles of a triangle is 180 °, the sum of ∠MHR and ∠MRH plus ∠HMR give π. So, since supplementary angles of congruent angles are congruent, and that ∠HMR is congruent to itself then the sum of ∠MHR and ∠MRH is congruent to ∠SMR, then ∠SMR > ∠MRH. If HS is the median of PR then ∠SMR = 2∠MRH, because MRH is an isosceles triangle so the angles at the base are congruent.

Bob · 2013-04-03 06:31:20

hi berliner,

Is this a test of how well I know my 'Elements'.

Actually I thought your first 'thesis' was original but whilst searching for the second part I stumbled upon book 1 proposition 16. This is a more general version of the thesis as it doesn't require a right angle. I am so sorry; your world fame must be delayed a little longer.

So then I turned to your second part:

If HS is the median of PR then ∠SMR = 2∠MRH, because MRH is an isosceles triangle so the angles at the base are congruent.

I could not see why MRH should be isosceles so I checked it out with an accurate diagram (thanks to Sketchpad) and you are right. But I think there's some steps missing from your proof. You will have to justify that the triangle IS isosceles.

In diagram 1 below, you can see that the triangle is isosceles as MP = MR = MH = radius of shown circle. But it only works if PHR = 90.

In my second diagram I have deliberately chosen PHR as much less than 90. Now the triangle MHR is not isosceles.

Keep up the good work

Bob

berliner · 2013-04-03 07:25:12

Hi Bob;

As I explained in the hypothesis PH is perpendicular to r, but don't worry. I'm trying to demonstrate that MRH is isosceles.

Bob · 2013-04-03 10:33:46

Yes I know you did. But the isosceles result only occurs under that condition, so you must use it to complete the proof. Without that angle the result doesn't follow.

I'll let you find out how. hint. I put the circle there for a reason.

Bob

berliner · 2013-04-04 06:00:19

hi Bob;

this is my demonstration but I dind't use the circle.

Hypothesis: P doesn't belong to r and H, R belong to r. M is the midpoint of PR.
Thesis: HM = MR

Begin to build the bisection s of ∠HMR, which is also the axis of simmetry of the angle. Since that s is the axis simmetry of ∠HMR:
Ss: HM <-> MR
Since that simmetry is an isometry and isometry manteins the distance between the points then HM = MR.
So, since that the triangle HMR has got two congruent line segments the triangle is isosceles.

Bob · 2013-04-04 06:56:05

hi berliner,

To show that this proof doesn't work I have deliberately not made PHR = 90.

Sketchpad assigns letters automatically so I couldn't stop it making H = A M = B M' = C and R = D

The triangle is not isosceles.

Now you may say, but the angle is 90 so what's wrong? Well you'll have to use that fact; otherwise my diagram is allowable.

Bob

berliner · 2013-04-04 07:11:36

Hi Bob;

It's impossible to demonstrate that the triangle is isosceles when PHR = 90. I never said that. I said that the triangle is isosceles only when PHR = 90.

Bob · 2013-04-04 07:45:51

hi Berliner

The triangle is isosceles. I'm not disputing that. My problem is I don't think you have proved it.

Let me give you a simple example to show what's wrong.

I'll make up some facts

(i) a, b and c are numbers.

(ii) a + b = 10

(iii) a = 7

(iv) c = 3

With those facts it is quite easy to prove that b = c.

But let's leave out a fact:

(i) a, b and c are numbers.

(ii) a + b = 10

(iii) c = 3

Can you now prove that b = c?

No you cannot. Fact (iii) is essential. It must be used in the proof.

PHR = 90 is like fact (iii). You cannot prove the isosceles result without it.

I have made your diagram using Sketchpad. The software is working with accuracy set to 2 dp.

With PHR = 90, however I move the points about, the triangle is always isosceles. If I take away the 90, it never is.

You have a triangle HMR and you bisect the angle at M. You cannot conclude from that alone that HM = MR. Otherwise every triangle would be isosceles. An axis of symmetry for an angle is not necessarily an axis of symmetry for the whole triangle.

I can prove it is an isosceles triangle. My proof uses the circle. I expect other methods are also possible. But they will have to use the 90.

Bob

berliner · 2013-04-04 08:09:04

Hi Bob;

I've an alternative demonstration:
Tracing the segment MM' which is perpendicular to r and M' is the midpoint of HR HM' = M'R, so. The simmetry of axis MM' transforms M in M (because axis points are united), M' <-> M' and H <-> R because HM' = M'R and M' = M'. Since that extrems corresponds then segments, HM and MR, must correspond and since they correspond they are congruent.

Bob · 2013-04-04 08:47:14

hi berliner,

You cannot have MM' perpedicular to r and M' is the midpoint of HR. M' may not have both these properties. (It does, but not based on what you have submitted)

I suggest this:

Draw a line parallel to PH through M. Let this line cut HR at point M'

As MM' is parallel to PH => HM'M = MM'R = 90. (parallels cuts by transversal HR)

And in triangle PHR, PM = MR => HM' = M'R ( our old friend EDIT book 6 proposition 2)

So consider the triangles HMM' and RMM'

HM'M = MM'R (above)
HM' = M'R (above)
MM' is common

=> HMM' and RMM' are congruent (SAS)

=> HM = RM => HMR is isosceles.

Note: I have used the 90 property.

Circle method:

With M as centre and PR as diameter draw the circle shown in the previous diagram.

Consider angle PHR = 90.

=> H must lie on the circumference of the circle (converse of 4th angle property of a circle, diameter subtends an angle of 90 at the circumference)

=> MH = radius = MR => HMR is isosceles.

Bob

berliner · 2013-04-04 09:17:18

Hi Bob;

Those are two brilliant methods. I said that M' is the midpoint of HR because it's the orthogonal projection of M. Perhaps there's an axiom which say that The orthogonal projection of the midpoint of a segment is the midpoint of the orthogonal projection of the segment itself.

If it doesn't exist we can try to demonstrate it.

Last edited by berliner (2013-04-04 09:17:59)

Bob · 2013-04-04 19:42:23

Once again, I refer to Euclid's Elements:

Book 6 Proposition 2

If some straight-line is drawn parallel to one of the sides of a triangle then it will cut the (other) sides of the triangle proportionally. And if (two of) the sides of a triangle are cut proportionally then the straight-line joining the cutting (points) will be parallel to the remaining side of the triangle.

This covers your case and does much more. Replace 'orthogonal projection' with 'any parallel line' and 'midpoint' with 'any proportionality'.

See diagram below, showing two parallels splitting two sides into thirds.

Bob

berliner · 2013-04-04 22:25:55

You're totally right, Bob. Well... I'll find something else

Bob · 2013-04-05 00:21:02

hi berliner

A geometry puzzle.

This isn't a new bit of geometry but it is certainly a problem that has bobbym and me stumped.

It arose at the end of a chapter for grade 8 students. The chapter is on Pythagoras theorem.

You will see as you read the thread that we both came up with the answer, but only by using maths far above the level of grade 8 and not using pythag. Once you have an answer, you can use pythag to check it; but how to get the asnwer in the first place, using maths at that level ?

So that's the challenge; solve it in a reasonably simple way that a grade 8 student could understand.

http://www.mathisfunforum.com/viewtopic … 82#p158982

That should set you thinking ....

Bob

berliner · 2013-04-06 02:20:50

Hi Bob;

Interesting problem, I found a solution, but it's a little bit complicated for grade 8 students. I'm trying to find an easier solution.

berliner · 2013-04-06 08:21:48

Hi Bob;

I used Erone's formula and I obtained this equation:

From this I obtained x=27.9374... and y: 32.3504...
It sounds like a strange result, but I can tell you that your solution is the best and more comprehensible and I'm a grade 8 student.

Last edited by berliner (2013-04-07 05:05:05)

berliner · 2013-04-08 04:39:23

Hi everyone;

This is the last theorem of this topic.

Hypothesis: P∉r and PH ⊥ r
H, A, R ∈ r and HA = HR
PM = MH and PM = MR
Thesis: MA > MA and MA ⊥ r
If MA ⊥ r we must demonstrate that ∠MAR = 90° and ∠ARM + ∠AMR = 90°. For hypothesis PH ⊥ r, so ∠PHA = 90° and ∠HPR + ∠HRP = 90°. ∠AMR = ∠HRP because they have in common the line r and PR. Triangles MAR and PHR are similar, because ∠AMR = HR ̂P and PR : MR = HR : AR since that HR = 2AR and PR = 2MR.
Since that MAR and PHR are similar all their angles are congruent, so ∠HPR = ∠AMR and ∠PHR =∠MAR and since that ∠PHR = 90° then MA = 90°, then MA ⊥ r.
In the first theorem of this topic we saw that distances from the midpoints of the oblique to the line are congruent. If we take PH as an oblique then MA = MH, because M is the midpoint of PR and M is the midpoint of PH. We know that MA > MH, because M is point which doesnt belong to r and MH is its distance to r and MA is an oblique and oblique carried out from the same point in a straight line, increasing the distance from the point to the line. So, MA > MH, MH = MA and MA > MA.

Last edited by berliner (2013-04-08 05:03:45)

Bob · 2013-04-08 19:23:58

hi berliner,

That result looks good to me, well done. But there's some things in your proof that may be typing errors. have another look at the bits I've highlighted in red.

Hypothesis: P∉r and PH ⊥ r H, A, R ∈ r and HA = HR PM = MH and PM = MR
Thesis: MA > MA and MA ⊥ r
If MA ⊥ r we must demonstrate that ∠MAR = 90° and ∠ARM + ∠AMR = 90°. For hypothesis PH ⊥ r, so ∠PHA = 90° and ∠HPR + ∠HRP = 90°. ∠AMR = ∠HRP because they have in common the line r and PR. Triangles MAR and PHR are similar, because ∠AMR = HR ̂P and PR : MR = HR : AR since that HR = 2AR and PR = 2MR.
Since that MAR and PHR are similar all their angles are congruent, so ∠HPR = ∠AMR and ∠PHR =∠MAR and since that ∠PHR = 90° then MAR = 90°, then MA ⊥ r.
In the first theorem of this topic we saw that distances from the midpoints of the oblique to the line are congruent. If we take PH as an oblique then MA = MH, because M is the midpoint of PR and M is the midpoint of PH. We know that MA > MH, because M is point which doesnt belong to r and MH is its distance to r and MA is an oblique and oblique carried out from the same point in a straight line, increasing the distance from the point to the line. So, MA > MH, MH = MA and MA > MA.

Bob

berliner · 2013-04-08 20:25:31

Hi Bob;

This is the correct version, sorry, but I write an entire word document I realized that copying the text on the forum did not keep the symbols then I had to rewrite all and I made some mistakes.

Hypothesis: P∉r and PH ⊥ r
H, A, R ∈ r and HA = AR
PM = MH and PM = MR
Thesis: MA > MA and MA ⊥ r
If MA ⊥ r we must demonstrate that ∠MAR = 90° and ∠ARM' + ∠AMR = 90°. For hypothesis PH ⊥ r, so ∠PHA = 90° and ∠HPR + ∠HRP = 90°. ∠ARM' = ∠HRP because they have in common the line r and PR. Triangles MAR and PHR are similar, because ∠ARM'= ∠HR ̂P and PR : MR = HR : AR since that HR = 2AR and PR = 2MR.
Since that MAR and PHR are similar all their angles are congruent, so ∠HPR = ∠AMR and ∠PHR =∠MAR and since that ∠PHR = 90° then MA = 90°, then MA ⊥ r.
In the first theorem of this topic we saw that distances from the midpoints of the oblique to the line are congruent. If we take PH as an oblique then MA = MH, because M is the midpoint of PR and M is the midpoint of PH. We know that MA > MH, because M is point which doesnt belong to r and MH is its distance to r and MA is an oblique and oblique carried out from the same point in a straight line, increasing the distance from the point to the line. So, MA > MH, MH = MA and MA > MA.

Math Is Fun Forum

#1 2013-04-02 01:01:27

Theorem in Euclidian geometry

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