perplexus.info :: Geometry : SSS, SAS, ASA, and ...

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.

Submitted by Bractals

Rating: 5.0000 (1 votes)

Solution: (Hide)

Let [B' | E'] be the foot of the perpendicular from [C | F] to
the extension of side [AB | DE].

   ∠B'BC = 180° - ∠ABC = 180° - ∠DEF = ∠E'EF
   ∴ ΔBB'C ≅ ΔEE'F by AAS
   ∴ |BB'| = |EE'| and |B'C| = |E'F|

   |AB'|² = |AC|² - |B'C|² = |DF|² - |E'F|² = |DE'|²
   ∴ |AB'| = |DE'|
   ∴ |AB| = |AB'| - |BB'| = |DE'| - |EE'| = |DE|
   ∴ ΔABC ≅ ΔDEF by SSS

QED

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
re(2): First part of solution	Tristan	2013-10-05 01:03:52
re: First part of solution	Bractals	2013-10-04 05:38:06
proof of lemma	Tristan	2013-10-03 06:04:31
First part of solution	Tristan	2013-10-03 05:45:34

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