In addition to not being drawn to scale and asking for congruence anyway, I like this problem because it potentially forces some great class discussions.

Sunday morning proof anyone? How would you tackle this? #mathchat http://t.co/v1TrbHSg1N—

Jim Noble (@teachmaths) January 25, 2015

One responder suggested using the Law of Sines (LoS) to establish an isosceles triangle. My first thought was that was way more sophisticated than necessary and completely missed the fact that the given triangle information was SSA.

My initial gut reaction was this SSA setup was a “trick” ambiguous case scenario and no congruence was possible, but I couldn’t find a flaw in the LoS logic. After all, LoS fails when attempting to find obtuse angles, but the geometry at play here clearly makes angles B and C both acute. That meant LoS should work, and this was actually a determinate SSA case, not ambiguous. I was stuck in a potential contradiction. I was also thinking with trigonometry–a far more potent tool than I suspected was necessary for this problem.

“Stuck” moments like this are GOLDEN for me in the classroom. I could imagine two primary student situations here. They either 1) got a quick “proof” without recognizing the potential ambiguity, or 2) didn’t have a clue how to proceed. There are many reasons why a student might get stuck here, all of which are worth naming and addressing in a public forum. How can we NAME and MOVE PAST situations that confuse us? Perhaps more importantly, how often do we actually recognize when we’re in the middle of something that is potentially slipperier than it appears to be on the surface?

PROBLEM RESOLUTION:

I read later that some invoked the angle bisector theorem, but I took a different path. I’m fond of a property I asked my geometry classes to prove last year .

If any two of a triangle’s 1) angle bisector, 2) altitude, and 3) median coincide, prove that the remaining segment does, too, and that whenever this happens, the triangle will be isosceles with its vertex at the bisected angle.

Once I recognized that the angle bisector of angle BAC was also the median to side BC, I knew the triangle was isosceles. The problem was solved without invoking any trigonometry or any similarity ratios.

Very nice problem with VERY RICH discussion potential. Thanks for the tweet, Mr. Noble.

For more conversation on this, check out this Facebook conversation.

For this triangle “angle bisector divides the opposite side in the ratio of adjacent sides” shows that AC = AB

Ah ha, I am glad it distracted you too! I have spent a good bit of time on this today – all fruitful. It is worth checking out this discussion thread on the Mathematics teacher exchange fb group! https://www.facebook.com/groups/182365128512083/766194596795797/?notif_t=group_comment

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