The Mathematics Education Department
EMT 725, Lisa F. Garey
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The problem asks for the measure of angle BCD. As the hint suggested, constructing a perpendicular line through the midpint of AD gives us AE = DE, where E is any point on that perpendicular line.
Let ADE be an equiateral triangle. 
Then we have 
.
The hint also suggested reflecting along CD. When ADC and their segements are reflected along CD, the image is conruent with EDC. Then we have a kite where the opposite angles are equal.
The interior angle sum of any quaddrilateral is 360. Let angle BCD be x. Hence,
360 = angle(DAC)+ angle(ADA') + angle(DA'C) + angle(A'CA)
= 100 + 60 + 100 + (40+x) + (40+x)
x = 10 degrees = angle(BCD)
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