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Yes I remembered now - my workings were with sin(100) rather than sin(80) but same thing obviously.Charles Link said:Very clever @neilparker62 :)
To show the triangles are similar, we need to show ##de/da=da/db ##. (These are corresponding sides around the same angle).
Now ## da/db=da/ac=\sin(40)/\sin(80) ##.
Meanwhile ## de/da=\sin(30)/\sin(50) ##.
We need to show ## \sin(40)/\sin(80)=\sin(30)/\sin(50)##.
With ## \sin(80)=2 \sin(40) \cos(40) ##, and ## \sin(30)=1/2 ##, and ## \cos(40)=\sin(50) ##, we have proven the triangles to be similar, and thereby angle ##B=30 ## degrees.