Triangle Midsegment Properties - Problem 2
In this problem we’re trying to find the variables a and b. If I look at this diagram, I see that a and b are angles and I see that I have a midpoint here because I can see that it’s dividing this segment into two congruent pieces, and the same can be true of this side as well.
So this is a midsegment and what do we know about midsegments? Well we know that a midsegment will be parallel to the third side, the side that is not an endpoint of the midsegment. So what I can do on this problem is say that this midsegment right here must be parallel to that third side, but why would that help us?
Well if I look at these two parallel lines, I see that I have a transversal which is one of those sides and I have another transversal here so I can use what I know about corresponding angles and same side interior angles to find these missing measures.
Let’s start off with a. I see that a corresponds to 80 degrees, they’re corresponding angles which means they must be congruent, so I’m going to write that a must be 80 degrees.
Second thing is finding b. B and 50 are on the same side of that transversal and they’re in between the 2 parallel lines, we call those same side interior. So b+50, if I wrote this out, must be supplementary which means they have to sum to 180 degrees.
So if you solve this for b, you’re going to subtract 50 degrees from both sides and you find that b must be 130 degrees. The key thing to this problem was realizing that we had a midsegment which we recognized from the markings that showed that we had congruence and remembering that the mid-segment is parallel to the third side.