The solution is by Ian McGee of the University of Waterloo.
Construct an isosceles triangle AEQ with ∠AQE = 20°. Since. AE = BC, the latter is equal to ΔABC. In particular, AQ = AC. Also,
∠CAQ = ∠EAQ - ∠EAC = 80° - 20° = 60°.
Which makes ΔACQ equilateral. In particular, CQ = EQ = AQ.
In ΔCQE, EQ = CQ and ∠CQE = 60° - 20° = 40°. Thus, ∠CEQ = (180° - 40°) / 2 = 70° so that ∠AEC = 80° + 70° = 150°.
Reference
- R. Honsberger, Mathematical Chestnuts from Around the World, MAA, 2001, Ch 2.
|Contact|
|Front page|
|Contents|
|Geometry|
|Up|
|Store|
Copyright © 1996-2012 Alexander Bogomolny