If a median in a triangle is also an angle bisector, then what
is the most straightforward method to prove that the triangle is
isoceles?

Note that I am not assuming the converse, namely that the triangle
is isosecles; I am given only the median and the angle bisector.
This gives two triangles in the interior that have an SSA relationship,
which cannot be used to prove them congruent.  So what is the proof?

Thanks--

Hi, whack-ga:

Here's one approach.  Extend the median from A through the side BC of
the triangle which it bisects at M an equal distance away from that
side to D, thereby creating a quadrilateral ABDC whose diagonals (the
extended median and the intersecting side of the original triangle)
are mutually bisecting (because the median bisected the original side,
and we extended the median across that side by an equal distance).  In
other words the midpoint M of BC is also the midpoint of AD.

Now we have the quadrilateral ABDC must be a parallelogram, by virtue
of its mutually bisecting diagonals.  The proof is given here:

[CA Parallelogram Diagonals]
http://www.math.washington.edu/~king/coursedir/m444a00/test/ca10-16-pgramdiag.html

but in brief the alternating angles at the intersection M give rise to
a proof of triangle congruence using SAS, which in turn gives us a
proof of the parallelism of opposite sides of the quadrilateral (via
the equality of alternate interior angles).  In particular triangle
ABM is congruent to triangle DCM.

Thus far we have used _only_ the property that AM is the median to BC,
and that is as far as it can carry us.  We now enlist the additional
fact that AM is the angle bisector at A, ie. that angle BAM = angle
CAM.  But angle BAM is also equal (by the conguence of triangle above
= alternate interior angles) to angle CDM.  Putting these together,
angle CAM = angle CDM.  Thus triangle CAD is an isoceles triangle with
base angles equal.

This forces the "legs" of CAD to be equal, AC = DC.  But DC was also
equal to AB, as the opposing side of the parallelogram (or going back
to the earlier triangle congruence between ABM and DCM, if you
prefer).  Thus AC = AB, which proves the original triangle ABC is
isoceles.  QED

If it helps, one can "summarize" what is really going on in this proof
by saying that the obstacle of the SSA "facts" that we are originally
given is overcome by splitting the triangle ABC along AM, and rotating
the ABM subtriangle around to join ACM along the equal sides BM and
CM.  D is in fact the new position attain by A after this rotation,
which accomplishes the desired goal by inverting the SSA data around
to become SAS data.  Naturally the construction depends heavily on the
fact that AM was the median drawn to BC.

regards, mathtalk-ga

---

Clarification of Answer by mathtalk-ga on 28 Oct 2003 19:43 PST

As a further insight, note that one way to prove a triangle with equal
base angles is isosceles (a result assumed in my proof above) is by
appealing to the ASA criteria for congruence.

For if the equal base angles be swapped, leaving the "base" side
between in correspondance to itself, then we have an ASA congruence of
the triangle to itself that forces the opposing sides to the base
angles to equal one another.

regards, mathtalk-ga

Thanks for the proof--I think it's easier for beginning
students to understand it by using the fact that the 
vertical angles AMB and DMC are congruent so that triangles 
AMB and DMC are congruent, CAM and CDM are congruent by 
transitivity, ACD is isosceles, etc. etc.

That way I don't need to introduce concepts of parallel 
lines, quadrilaterals, etc. and can just focus on triangle 
theorems.

Thanks a bunch, mathtalk-ga!