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 Subject: Isosoles Triangle proof Category: Science > Math Asked by: whack-ga List Price: \$3.50 Posted: 28 Oct 2003 14:56 PST Expires: 27 Nov 2003 14:56 PST Question ID: 270589
 ```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```
 whack-ga rated this answer: ```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!```