Thank you so much. It's so helpful to see the propositions laid out so that one might construe the implications cohesively (as opposed to point-by-point) and then reconstruct them.
At 1:48 you used proposition 6 to show that since 2 sides are equal, the angles are equal. While I know that this is true, proposition 6 only said the reverse, that if 2 angles are equal then the sides are equal. I found this to be a curiosity.
What if the points C and D are at the same vertical distance from the line AB and symmetrical with respect of it's mid point? (the shape would be almost like a "W" turned upside down). In this case i think the two lines wold be equal but i'm not able to proof or disproof this statement.
If that were the case, the two triangle that you create would be similar but flipped, all sides equal but with the RIGHT side of the original triangle would be equal to the length of the LEFT side of the new triangle, and vice versa This proposition is very specific that the length of the of RIGHT side of the triangle does not change.
Or what if point D is under line AB, so that it is equidistant to the line as point C, perpendicular to line AB? Or, if you need it flipped for the sake of being chiral, on the bottom and mirrored along the half way point of AB?
Thank you so much. It's so helpful to see the propositions laid out so that one might construe the implications cohesively (as opposed to point-by-point) and then reconstruct them.
+Rachel Goad Thank you
It was so helpful, thanks a lot
Nice and enjoyable.
What if they meet at a point inside the triangle? There's more than one case to prove.
Great presentation series BTW. Thanks for making it.
At 1:48 you used proposition 6 to show that since 2 sides are equal, the angles are equal. While I know that this is true, proposition 6 only said the reverse, that if 2 angles are equal then the sides are equal. I found this to be a curiosity.
I made a mistake, it was supposed to be proposition 5 "In an isosceles triangle, the angles at the base are equal to one another"
@@SandyBultena Ah, thank you for the correction and clarification. Kickass!
Should this proof go before proposition 4 (SAS)? Thoughts?
No, because it relies on I.5, which in turn relies on I.4. This proof is useful as a lemma for I.8 (SSS congruence).
this is only true for one side of the line right? there is a point below (basically mirror image) that would be another intersection
Yes you are correct
How are the angles equal please? Thank you
An isoceles triangle has equal angles at the base. You can watch Proposition 6 for a proof if you need it.
What if the points C and D are at the same vertical distance from the line AB and symmetrical with respect of it's mid point? (the shape would be almost like a "W" turned upside down). In this case i think the two lines wold be equal but i'm not able to proof or disproof this statement.
If that were the case, the two triangle that you create would be similar but flipped, all sides equal but with the RIGHT side of the original triangle would be equal to the length of the LEFT side of the new triangle, and vice versa
This proposition is very specific that the length of the of RIGHT side of the triangle does not change.
Or what if point D is under line AB, so that it is equidistant to the line as point C, perpendicular to line AB? Or, if you need it flipped for the sake of being chiral, on the bottom and mirrored along the half way point of AB?
The proof doesn't work when point D is inside triangle ABC.
"Unique" is not in the definitions.
Easy on the formalism buddy.