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SSS, SAS, ASA, and ... (Posted on 2013-10-02) Difficulty: 4 of 5

Let ABC and DEF be triangles with

    ∠ABC = ∠DEF > 90°,
    |BC| = |EF|, and
    |CA| = |FD|.

Prove that ΔABC ≅ ΔDEF with a direct geometric proof.


See The Solution Submitted by Bractals    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: First part of solution | Comment 3 of 4 |
(In reply to First part of solution by Tristan)

Hi Tristan. Glad to see you back.


I liked your proof, but don't think you need the lemma.
The Pythagorean  theorem gives the third side.

You have probably seen the proof of the Steiner-Lehmus theorem using the above theorem. Since your proof of this theorem is direct and geometric it would seem that the proof of the Steiner-Lehmus theorem is also direct and geometric. But, John Conway does not think one exists. What do you think?

Edited on October 5, 2013, 5:03 pm
  Posted by Bractals on 2013-10-04 05:38:06

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