Let ABC and DEF be triangles with

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

Prove that ΔABC ≅ ΔDEF with a direct geometric proof.

Previously, I showed that it is sufficient to show for the case where
∠ABC = ∠DEF = 90°

Start by dropping a perpendicular from B to AC.  Let the intersection with AC be point P.  Similarly, construct point Q on the triangle DEF.  This is a standard geometrical construction where you divide a right triangle into two smaller right triangles, each similar to the larger one.

By similarity, |BC|/|AC| = |PC|/|BC|
so |AP| = |AC| - |PC| = |AC| - |BC|²/|AC|
Using the same argument, we can show that |DQ| = |DF| - |QF| = |DF| - |EF|²/|DF|
Therefore |AP| = |DQ|

By similarity, |AP|/|AB| = |AB|/|AC|
So |AB| = sqrt(|AP|*|AC|)
Using the same argument, |DE| = sqrt(|DQ|*|DF|)
So |AB| = |DE|

This is sufficient to prove congruence by SSS or SAS.

Posted by Tristan on 2013-10-03 06:04:31