In an acute triangle ABC, the point D is a foot of an altitude from vertex C. Points E and F lay respectively on sides AC and BC, while AE=AD and BF=BD. Point S is symmetrical to point C against the middle of the circle circumscribed on the triangle ABC. Show that SE=SF.
Could you help me with this? Should I use the law of sines? I can't manage to get through it...
September 7th 2011, 09:49 AM
Re: Showing being even
I would feel a whole lot more inclined to think about this if you were to upload a reasonable sketch of the given conditions.
Heck, YOU might even solve it with a good enough picture! ;)