Friday Puzzler -- Angle

Re: Friday Puzzler -- Angle

Larry Barrett

Label the vertices thusly: A is the lower left hand vertex. Then, moving CCW, B is the lower right hand vertex, then C and D. Label the intersection in the middle E. You can also extend the two long sides to their intersection, which is F.

It is easy to show that triangle AED is isoceles with angle ADE and angle AED = 75*. Triangle BCE is also isoceles with angle BEC= angle BCE=75*. And angle AEB = angle CED=105*.

Also, angle AFB=45*, and angle FDE = angle FCE=105*.

This is as far as I can go. I believe that triangle DCE is similar to triangle AEC and therefore the angle in question, angle DCE is 44*, but can't prove it yet.

On the right track

Alex Y

toward the right conclusion!

Final Step

Alex Y

Larry got this one, but here is the final step.

First, look at the angles Larry got, basically from sum of angles of a triangle = 180 and 180* in a straight line.

Triangle DEC is similar to AEB if an angle is identical (in this case, 105*), and the edges on either side of that angle are the same multiple of the edges on the other triangle. The sides in question on AEB are the long sides of the two similar isosceles triangles, and the sides of DEC are the bases of those two triangles. Obviously, the bases of those two isosceles triangles are the same multiple of their other two sides.

That doesn't seem to go very well in words--maybe I will edit tomorrow morning after a mug of coffee!