100 Degree Isoceles Triangles

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Though innocent looking - this problem can keep you busy for a very long time and then when you see a solution it seems almost trivial.....


Here goes:

Construct DE so that angle BDE = 60 degrees

Since BDE = 60 deg. is the vertical angle of an isosceles triangle, it follows that triangle BDE is in fact equilateral.

Hence angle ABF = 40 deg. and it follows that triangle ABF is isosceles with BF = FA and as a result FC = FE

By the SAS case of congruency it follows that triangle BFC is congruent to triangle AFE and it follows that AB = AE.


It now follows that ABDE is a kite and AD intersects BE at 90 deg.


And it follows that the elusive angle DAC = 10 deg.