Proof using Complex Numbers: Ptolemy's Theorem

We prove Ptolemy's theorem, and its converse, using complex numbers. More precisely, we show that for a quadrilateral ABCD, |AB||CD| + |BC||AD| = |AC||BD| holds if and only if ABCD is a cyclic quadrilateral.
A quadrilateral is cyclic if and only if opposite angles sum to 180°:
en.wikipedia.org/wiki/Cyclic_...
The proof here is inspired by Hahn's proof in this book:
Hahn, L., 1994. Complex numbers and geometry. Cambridge University Press.
00:00 Ptolemy's theorem
01:06 Ptolemy's inequality
03:56 Equality in Ptolemy's inequality
06:21 Strategy for the rest of the proof
07:13 Equality if and only if...
09:35 Relating arguments to angles
12:34 Conclusion

Пікірлер: 14

  • @chengkaigoh5101
    @chengkaigoh5101 Жыл бұрын

    Complex analysis is so interesting

  • @renesperb
    @renesperb Жыл бұрын

    A nice application , showing how useful complex numbers can be for treating geometric problems .

  • @squeezy8414
    @squeezy8414 Жыл бұрын

    Just did this in a Cambridge STEP question, really cool stuff!

  • @KSignalEingang
    @KSignalEingang Жыл бұрын

    [Bulgarian accent] Verrry sophisticated. (If you know, you know)

  • @josephb5417

    @josephb5417

    Жыл бұрын

    I know, love zvezdelina

  • @wannabeactuary01
    @wannabeactuary01 Жыл бұрын

    At 5:38 Let z = Rexp(im) and w = Texp(im) where R and T are moduli and m is the common argument of both complex numbers z/w = R/T = (R/T) exp(0) ie argument is zero as number is real and modulus R/T It might be obvious but never ever recall encountering this before!

  • @DrBarker

    @DrBarker

    Жыл бұрын

    Not only is z/w real, but also z/w is a positive real number when they have the same argument. And z/w is negative iff arg(z) = -arg(w), so they lie on opposite sides of the same line through the origin.

  • @user-rp6yq5rw3r
    @user-rp6yq5rw3r Жыл бұрын

    Amazing bro!

  • @avyakthaachar2.718
    @avyakthaachar2.718 Жыл бұрын

    So neat👌👌

  • @t.sambath
    @t.sambath Жыл бұрын

    Love ❣️❣️

  • @user-gs4cq3uk2b
    @user-gs4cq3uk2b9 ай бұрын

    I had never seen a vector proof of Ptolemy's theorem, so I tried it myself. (Proof by Vector) Let the position vectors of points B, C, and D be b, c, and d, respectively, with A as the starting point. AB・CD+AD・BC=AC・BD ⇔ |c-d||b|+|c-b||d|=|b-d||c| ① ∠ACD=∠ABD ⇒ c・(c-d)/(|c||c-d|)=b・(b-d)/(|b||b-d|) ② ∠ADB=∠ACB ⇒ d・(d-b)/(|d||d-b|)=c・(c-b)/(|c||c-b|) ③ ∠ADC+∠ABC=180° ⇒ d・(d-c)/(|d||d-c|)+b・(b-c)/(|b||b-c|)=0 ④ left side of ① = |b|(c-d)・(c-d)/|c-d|+|d|(c-b)・(c-b)/|c-b| =|b|((c-d)・c/|c-d|-(c-d)・d/|c-d|)+|d|((c-b)・c/|c-b|-(c-b)・b/|c-b|) (substitute ②④③ into the 1st, 2nd, and 3rd terms, respectively) =|b|((|c|/|b|)b・(b-d)/|b-d|-(|d|/|b|)b・(b-c)/|b-c|)+|d|((|c|/|d|)d・(d-b)/|d-b|-(c-b)・b/|c-b|) =|c|(b-d)・(b-d)/|b-d| =|c||b-d| =right side of ①

  • @DrBarker

    @DrBarker

    9 ай бұрын

    Very neat, there are lots of ways to prove Ptolemy's theorem!

  • @VanNguyen-kx6gx
    @VanNguyen-kx6gx Жыл бұрын

    Certified very long and wasted time in geometry. Use vectors to verify easily.

  • @TechToppers

    @TechToppers

    3 ай бұрын

    I don't even think he went over a lot of geometry in the video aside from the fact that opposite angles are supplementary iff the quadrilateral is cyclic...