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Weekly Problem 18 - 2011

Stage: 2 and 3 Short Challenge Level: Challenge Level:1

Since ABCD is a square, $\angle BCD = 90^{\circ}$,

and since CDE is an equilateral triangle, $\angle DCE = 60 ^{\circ}$.
 
Thus $\angle BCE = \angle BCD + \angle DCE = 90^{\circ}+60^{\circ}=150^{\circ}$.
 
Because CDE is an equilateral triangle, $EC = DC$ and also, because ABCD is a square, $DC = CB$. Hence $EC = CB$ and ECB is an isosceles triangle.
 
So $\angle CEB = \angle CBE = \frac{1}{2} (180-150)^{\circ} = 15^{\circ}$, and hence $\angle BED = \angle CED-\angle CEB = 60^{\circ} - 15^{\circ} = 45 ^{\circ}$.
 
 
 
 
 

This problem is taken from the UKMT Mathematical Challenges.

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