@PCGamerPortal "first semester" which? U of Chicago undergrad and UNC grad never seen. Most programs assume that you will acquire this sort of knowledge recreationally.
@AtheistCitizen my math programme at the National Autonomous University of Mexico includes a first semester course entitled modern geometry where we learn about euclid's axiomatic system and construct our geometry from them, later adding other axioms like pasch's axiom. I think it is a very complete course which would be hard to top recreationally by most aspiring mathematicians.
@PattyManatty Because the area of DAB = 1/2 * area of CEDA and area of CAH = 1/2 * area of AHIK , this is because DAB has the same base and height as CEDA, same for AHIK and CAH.
Thanks. Well done. I remember reading about this specific proof earlier this year, but your illustrated video proof is a lot simpler, and reinforces the proof visually. It is a great version of this proof, though I remember when I first saw this proof, I was unsatisfied -- can't remember why, cuz it seems pretty complete to me : )
since the formula obviously works for right triangles, we can think of the obtuse triangle as a big right triangle (with height same as the obtuse and base equal to the extension of the base to the height PLUS the original base) minus a small right triangle (height same as the obtuse and base equal to the extension of the base to the height). draw the diagram and you'll understand.
@anticorncob6 you prove that the area of a paralelogram is bh, then any 2 congruent triangles can be arranged into a paralelogram with the same base and height as the triangles.
@ToddsGen This proof is better then the others because the others just prove that a^2 + b^2 = c^2, where this proof shows that along with that the area of ECAD is the same as AKIH (see video for what I mean), and the same with the other two.
There might be a simpler proof of this, but this is an original. It's cool and should be appriciated.
Today I want to show you how to use a mic
tibschris 3 months ago 4
you need a better microphone
PCGamerPortal 4 months ago
No need to do the 2nd part!!! As a math major, even grad student, I never spent any energy on such a proof, hence I must thank you for your pf
AtheistCitizen 4 months ago
@AtheistCitizen that's weird I saw several proofs of this theorem in my first semester
PCGamerPortal 4 months ago
@PCGamerPortal "first semester" which? U of Chicago undergrad and UNC grad never seen. Most programs assume that you will acquire this sort of knowledge recreationally.
AtheistCitizen 3 months ago
@AtheistCitizen my math programme at the National Autonomous University of Mexico includes a first semester course entitled modern geometry where we learn about euclid's axiomatic system and construct our geometry from them, later adding other axioms like pasch's axiom. I think it is a very complete course which would be hard to top recreationally by most aspiring mathematicians.
PCGamerPortal 3 months ago
what program are you using?
MichaelDamoulianos 6 months ago
I don't understand why DAB, and CAH having the same area implies that AHIK and CEDA have the same area?
PattyManatty 7 months ago
@PattyManatty Because the area of DAB = 1/2 * area of CEDA and area of CAH = 1/2 * area of AHIK , this is because DAB has the same base and height as CEDA, same for AHIK and CAH.
PCGamerPortal 4 months ago
Thanks. Well done. I remember reading about this specific proof earlier this year, but your illustrated video proof is a lot simpler, and reinforces the proof visually. It is a great version of this proof, though I remember when I first saw this proof, I was unsatisfied -- can't remember why, cuz it seems pretty complete to me : )
ThisIsRiyad 9 months ago
@anticorncob6 the proof is simple to understand.
since the formula obviously works for right triangles, we can think of the obtuse triangle as a big right triangle (with height same as the obtuse and base equal to the extension of the base to the height PLUS the original base) minus a small right triangle (height same as the obtuse and base equal to the extension of the base to the height). draw the diagram and you'll understand.
bryanwangpengjun 11 months ago
much easier to understand compared to the wikipedia article. Thankyou.
wullz16 1 year ago
Prove the formula bh/2 works for obtuse triangles. I'm sure that's right but I want proof.
anticorncob6 1 year ago
@anticorncob6 you prove that the area of a paralelogram is bh, then any 2 congruent triangles can be arranged into a paralelogram with the same base and height as the triangles.
PCGamerPortal 3 months ago
@PCGamerPortal Thanks for responding even though my comment was nine months old. I think I understand it.
anticorncob6 3 months ago
@anticorncob6 lol I didn't notice that xD
PCGamerPortal 3 months ago
lol, this is over complicated... why make it harder then it is? that is a silly thing to do.
ToddsGen 1 year ago
@ToddsGen This proof is better then the others because the others just prove that a^2 + b^2 = c^2, where this proof shows that along with that the area of ECAD is the same as AKIH (see video for what I mean), and the same with the other two.
There might be a simpler proof of this, but this is an original. It's cool and should be appriciated.
anticorncob6 1 year ago
amazing! this is a great proof, thanks a lot!
mrtamborineman10 1 year ago
@nymathteacher thanks u so much it helps alot
dalis001 1 year ago
how can i prove those four triangles half of the square??
dalis001 1 year ago