I started by looking at problem 45 on gogeometry.

My first instinct here was to increase the overall symmetry in the drawing by reflecting everything.

Once that's done through a variety of different angle chases you repeatedly arrive at 2b + 2a + x = 90.

The easiest place to do this is in corner BAC once the perpendicular bisector is drawn.

I could see the parallel lines in my reflected drawing which gave me a few more x angles but not another triangle equation I could use to solve the problem.

I next tried a few unsuccessful techniques. I back-cracked from the answer 30 degrees and filled in the other angles to see if any other common triangles were there that I had missed on visual inspection.

I then removed all the x's and replaced them with 90 - 2b - 2a in the total picture to see if this helped.

I looked around to see if there was any way to use the 30-60-90 properties of side length to deduce the angle measure but that looked unlikely given the constraints.

Then I took a break for a day after hitting a wall. And

*suddenly*on restarting during lunch I noticed that I had the in-center of the triangle when redrawing the picture of one side. That meant the third side was also an angle bisector and everything fell into place. [**The need for a break before finding a solution on a stumper is pretty normal for me]**
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