Math Is Fun Forum

Al-Allo

Member

Registered: 2012-08-23

Posts: 324

To a given straight line in a given rectilinear angle, to apply a ...

Sorry Bob, I hope I'm not bothering you with my numerous questions.

There's again one small detail on which I'm not sure. (Proposition 44 - book 1)

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI44.html

Here's the quote :

"Then HLKF is a parallelogram, HK is its diameter, and AG and ME are parallelograms, and LB and BF are the so-called complements about HK. Therefore LB equals BF."

So, we have constructed a parallelogram HLKF and we have also constructed the diameter HK which bisects it. Now, the only part which I'm not sure is concerning the small parallelograms AG ME and the complements LB BF. We know that by propositon 43, the complements of a parallelogram equal one another. My question is : knowing that I have parallelogram that has a bisector, do we automatically know that there are two parts which are smaller parallelograms and that there are two other parts which are complements ? Can we just assume that these four smaller parts exists without proving it in proposition 44 ? Is saying "ok, here are two complements and two smaller parallelograms) enough ?

Thank you!

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: To a given straight line in a given rectilinear angle, to apply a ...

hi Al-Allo

It's taken me a while to read all the props that leads up to this one.  I'm not sure I fully understand your query.  The lines from a parallelogram are extended so this creates several parallelograms and allows the earlier props to be applied.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Al-Allo

Member

Registered: 2012-08-23

Posts: 324

Re: To a given straight line in a given rectilinear angle, to apply a ...

hi Al-Allo

It's taken me a while to read all the props that leads up to this one.  I'm not sure I fully understand your query.  The lines from a parallelogram are extended so this creates several parallelograms and allows the earlier props to be applied.

Bob

What I mean is, shouldn't I prove that we have two small parallelograms ? That we are in fact dealing with such objects ? (Which I Would need to prove) By the way, I finished book 1 of the Elements. A pretty good book and interesting The only irritating part is when you have to draw many many many things.

Last edited by Al-Allo (2015-05-24 18:12:13)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: To a given straight line in a given rectilinear angle, to apply a ...

hi Al-Allo

Euclid was very  thorough, so I doubt that he left this out.  I don't know the propositions well enough to say where, but I'm sure he has already done this, in an earlier proposition.  Why not do it anyway as an exercise ?

I find I cannot do geometry unless I draw the diagrams.  My brain seems to be very visual;  I cannot remember names unless I see them written down.

Well done for completing book 1.    Only another 12 to go ...

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Al-Allo

Member

Registered: 2012-08-23

Posts: 324

Re: To a given straight line in a given rectilinear angle, to apply a ...

hi Al-Allo

Euclid was very  thorough, so I doubt that he left this out.  I don't know the propositions well enough to say where, but I'm sure he has already done this, in an earlier proposition.  Why not do it anyway as an exercise ?

I find I cannot do geometry unless I draw the diagrams.  My brain seems to be very visual;  I cannot remember names unless I see them written down.

Well done for completing book 1.    Only another 12 to go ...

Bob

I'm able to prove that the ABGH and BEFG are parallelograms. BEFG is one because we constructed it. ABGH is one because we can easily prove that we have 2 pairs of parallels. We have AB that is in a straight line with BE. And we have BE that is parallel to GF. Logically, AE is parallel to GF. We extend GF in a straight line to H. This makes AE parallel to HF. Finally, this makes AB parallel to HG (If we "cut" the bigger parallels) and AH is parallel to BG because we made the construction. I need to find a way to prove that MKEB has two pairs of parallels (thus, making it also a parallelogram) and this will result in proving that the fourth figure is really a parallelogram.

By the way, is my explanation any good you think ?

Thanks! (I do hope Im not bothering you in any way)

Last edited by Al-Allo (2015-05-25 06:14:49)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: To a given straight line in a given rectilinear angle, to apply a ...

hi,

In his proof he proves that HLKF is a parallelogram.  You can use that.

(I do hope Im not bothering you in any way)

I enjoy helping people on MIF.  I'm reasonably good at geometry but I've never gone through all the propositions like you have.  That makes it harder because I have to look them up and work through several just to understand what has already been proved.  My version of Java blocks the diagrams from that site as it says it is 'untrusted'.  But I have a pdf with the complete Elements which uses the same words and diagrams so I can get what I need from there.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob