SSA Proof

Date: 03/14/2001 at 14:54:05
From: Brent Gillett
Subject: Re: Geometry SSA congruence

How do I prove the hypotenuse-leg congruence for triangles? My 
professor suggested using the S.S.A. theorem

Date: 03/14/2001 at 17:00:41
From: Doctor Rob
Subject: Re: Geometry SSA congruence

Thanks for writing to Ask Dr. Math, Brent.

If two triangles have one angle, one adjacent side, and the opposite 
side equal, then they may or may not be congruent. That means that 
using S.S.A. is tricky.

Suppose, for example, you were given acute <A and sides b and a.  Then 
the picture might look like this:

                             C _,-'
                           _,o'
                       _,-'  .
                 b _,-'      .
               _,-'         d.
           _,-'              .
       _,-'                  .
   A o-----------------------o-----------
                             X

If you know some trigonometry, you'll realize that d = b*sin(<A). Now
draw a circle with radius a and center C. Point B must lie on line
AX and on the circle. There are several cases:

Case 1:  a < d.  The circle and line don't intersect, and there are no
triangles with these parameters.

Case 2:  a = d.  The circle and line are tangent at X, so B = X, and 
there is just one triangle with these parameters. In this case, <ABC 
is a right angle.

Case 3:  d < a < b.  The circle and line intersect in two points B and 
B', with B' between A and X, and B to the right of X. Now there are 
two triangles with these parameters, ABC and AB'C, and you can't prove
congruence.

Case 4:  b <= a.  The circle and the line intersect in two points B 
and B', but B' is at or to the left of A, so the angle drawn above is 
an exterior angle to that triangle. Thus there is just one triangle 
with these parameters, and you can prove congruence.

If <A is a right angle, and so d = b, the following cases hold:

Case 5: a <= b. No triangle.

Case 6: b < a. There is just one triangle, and you can prove 
congruence.

If <A is obtuse, then following cases hold:

Case 7: a <= b. No triangle.

Case 8: b < a. There is just one triangle, and you can prove 
congruence.

Thus the theorem should say,

THEOREM (S.S.A.):  If two triangles ABC and DEF have BC congruent to 
EF, AC congruent to DF, and <A congruent to <D, then:

  1. if AC >= BC, the triangles are congruent.

  2. if BC = AC*sin(<A), then the triangles are congruent.

  3. if AC*sin(<A) < BC < AC, then the triangles may or may not be
     congruent.

In your case, you know that the hypotenuse AC is longer than either 
leg, including BC, so Part 1 of the above theorem applies.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/