Euclid's Elements Book 1: Proposition 7, Side Side Side Theorem 1
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- Опубліковано 22 бер 2017
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This is the seventh proposition in Euclid's first book of The Elements. This proof focuses on the basic idea of the Side Side Side (S.S.S.) theorem.
You teach better than my teacher many thanks ;)
I'm glad that I could help. Thank you for watching.
This was very helpful. My teacher can’t taech and is boring, this helps a lot
Thank you for watching and for the kind words, I appreciate it.
Thank you
Thank you for watching. I hope the video was helpful.
Once again you save my life. Thank you so much
Hi Tommy
It's a small world.
guess so
Wow, even Bob Ross is here
Hello
Thank goodness, it's midnight and I have three propositions due tomorrow T-T
Bro sameee
2:13 I think you meant Proposition 6
"If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another."
Which is the same thing as saying,
'"If in a triangle two sides be equal to one another, the angles which subtend the equal sides will also be equal to one another."
Which is basically what you said.
love the colors btw
He was actually talking about proposition 4, which is that if two sides are equal to each other in a triangle, then the those two angles must be equal to each other (ex: in triangle ABC [A on left, B on top, and C on right] if line AB equals line AC then angle BAC must equal angle BCA)
Also there is interesting a new point in a way I call f created as by product ie we get extra value out of it or do we lose it? Not sure
How do we know CB and DB are equal? (forming an isosceles triangle)
Thanks for watching. That's a great question. Basically, in the proof by contradiction, we make assumptions and then show that the assumptions lead to a logical contradiction. This then proves that the original assumption must not be true.
For this proof, we start by assuming that these three unique lines can form a different triangle than the one we start with. We can assume that the lines coming from A and B do not meet at C but instead meet at a point D, thus forming a different triangle. Keep in mind though that we are not changing the lengths of the original lines. Therefore, the line coming from B and going to this new point D (line DB) would be equal to line CB. This is all part of the original assumption. Now, as seen in the proof, this assumption leads to contradictions and cannot be true.
You can try this on your own. Start with 3 different length sticks or pencils or items that are straight. Only one possible triangle can be made with these 3 unique lengths which is all this proposition is trying to show.
I hope this helps, but feel free to ask follow up questions.
@@EulersAcademy thank you so much!!
@@91JLovesDisneyNo worries, I'm glad I could help.
how do we know AC and AD are equal to each other?
I was going to ask the same question...help please
In the given of the problem, we already know that there are two straight lines that meet at a point. Additionally, we know that the statement we are trying to prove is that there are not two lines on the same side of the initial line (I.e. the line connecting A and B ) that meet in a different point and are EQUAL to the first two lines respectively. Therefore, when we assume for the sake of contradiction that our conclusion is actually false, we are actually assuming that there ARE two lines on the same side of the initial line that meet at a different point AND are Equal to each other respectively. This means when we start our construction we can already assume that the two lines we construct are equal to each other respectively. (i.e. this means that we assume that AC =AD and BC=BD)
Hope this helps!
Hello, cannot see that angles acd are the same as adc one looks much bigger ie obtuse the other is acute ? Thanks
You shouldn't go by drawings in proofs in general, you should go by the logic. AC=AD, that's a given, and because of that we have an isosceles triangle ACD, and an earlier proposition that the angles opposite the equal sides in an isosceles triangle are equal was proven.
how old are the students learning this?
Just curious
Honestly, it's a wide range of ages. Some people learn this on their own at a young age, some people learn this during their high school geometry class, some learn Euclid in college, and many others decide to learn the Elements as adults.
In my class were supposed to have only 9th but we have a senior and a scattering of 8th including me.
THANK U!
Hlo
Euclid's argument is incomplete. Angle DCB is only less than angle ACD when D is to the right of C, and it's only visually obvious (which seems to be the justification for this step) when D is placed where it is.
Thank you for watching. I will consider this point and possibly remake the video sometime later this year.
The (1862) edition I have also gives the case for if D is inside triangle ABC. You extend AC and AD as rays AE and AF from A, and use the exterior angle congruence of I.5 to build the contradiction.
@@35USCode Thank you for letting me know. I will certainly keep this point in consideration if I remake the video.
The proof doesn't work if point D lies inside triangle ABC.
It’s like 120 degrees vs 35 ish
Pretty sure that acd adc angles are the same it’s just the way it looks that confuses me. Wouldn’t minds clarification still.
@@timothyhill1149 You're correct, angle ACD is obtuse and ADC is acute in the drawing.
Since this is a proof by contradiction, you first assume something false (in this case, that D =/= C) and then use sound, logical deductions (in the case of Euclid's Elements, previously proven propositions, definitions, postulates, or common notions) to conclude something absurd. Then, since each deduction was sound and you arrived at a contradiction, the only explanation is that your original assumption was incorrect.
In this proof, it's given that AC = AD. Then, according to proposition 5 in book 1 (base angles in an isosceles triangle are equal), angle ACD = angle ADC. However, in the drawing, angle ACD =/= angle ADC as you noted. This is because Euclid intentionally makes a false assumption for the purpose of proof by contradiction that C =/= D. Because of where D is placed in the drawing, AC is visually shorter than AD (violating the given condition that AC = AD) causing the angles to be different. (Proposition 6 in book 1 says that if two angles in a triangle equal each other, then the sides across from those angles also equal each other. Hence, by contraposition, if AC =/= AD, then angle ACD =/= angle ADC.)
Hope that helps.
no u