Proof of Ptolemy’s Theorem

A recap of the proof of Ptolemy’s theorem I did in class today. This is a well known proof.   Simple and constructive in nature.

ptolemy

Given cyclic quadrilateral ABCD as shown. From point C, make line CE such that ∠ECD ≌ ∠BCA.

One can easily see that

ΔABC ∽ ΔDEC

which implies AB : DE = AC : DC … (1)

Also because ΔCAD ∽ ΔCBE,
which implies BC : AC = BE : AD … (2)

By (1) :   AB * DC = AC * DE

By (2) :  BC * AD = AC * BE     Adding these two equations gives

AB * DC + BC * AD = AC * DE + AC * BE = AC * (DE + BE) = AC * DB    Voila.

5 Comments »

  1. dedusuiu said

    as is

  2. dedusuiu said

    also we see that as say prof dr mircea orasanu this aspect is fundamental

  3. gocionu said

    with

  4. cedesu said

    important theorems obtained by prof dr Mircea Orasanu and prof drd horia orasanu as can approached followed using considerations and theorems of Girard Desargues special in cases of elliptic , hyperbolic and parabolic situations

  5. dudiasu said

    in many problems can be observed that theorems of Desargues considered prof dr nircea orasanu and prof drd horia orasanu can be applied to other domains as Constraints Optimizations with non holonomic aspects and prof dr Gigel Militaru with his doctoral students appear and thus these lead to Adrien Legendre theorems and results

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