Let ABC and DEF be triangles with

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

Prove that ΔABC ≅ ΔDEF with a direct geometric proof.

My proof has two parts.  First, I reduce it to a problem where ∠ABC = ∠DEF = 90.  Second, I prove it for the right angle case.

First, drop a perpendicular from C to AB.  Since ABC is obtuse, it will be necessary to extend AB to a new point B'.  Similarly, construct point E'.

I want to show that ΔBB'C is congruent to ΔEE'F.

∠CBB' = 180 - ∠ABC = 180 - ∠DEF = ∠FEE'
∠BB'C = ∠EE'F = 90
and |BC| = |EF|
By AAS, ΔBB'C is congruent to ΔEE'F

Now we know that |B'C| = |E'F|, and ∠AB'C = ∠DE'F.  If we replace B with B' and E with E', this is identical to the original problem, but with a right angle.

So let's call the right angle case a lemma.  The right-ASS lemma, if you will.

By the right-ASS lemma, |AB'| = |DE'|

Therefore |AB| = |DE|
Then we can prove the theorem using SAS or SSS.

Posted by Tristan on 2013-10-03 05:45:34