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 SSS, SAS, ASA, and ... (Posted on 2013-10-02)

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 No Rating

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 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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