In the square ABCD, E is the midpoint of AB. A line is drawn from E, perpendicular from CE, and intersecting AD at F. Find the ratio of angle BCF to angle BCE.
+128474 tan (BCE) = EB/ BC = (1/2)S/ S = (1/2)
So arctan (1/2) = angle BCE
EC = sqrt (S^2 + (1/2S)^2 ] = (S/2)sqrt (5) = [ sqrt (5) / 2 ] S
And angle EEF = angle BCE ≈ 26.565° = arctan (1/2)
So
cos (artan (1/2) ) = (1/2S) / EF
2 / sqrt (5) = S/ (2EF)
2EF / S = sqrt (5) /2
EF = [ sqrt (5) / 4 ] S
So tan ECF = EF/ EC = [ sqrt (5) / 4 ] S 1
_____________ = ___
[sqrt (5 )/ 2 ] S 2
So arctan (1/2) = ECF
So angle BCF = angle BCE + angle ECF = arctan (1/2) +arctan (1/2)
And angle BCE = arctan (1/2)
So
Angle BCF [ arctan (1/2) + arctan (1/2) ] 2arctan (1/2) 2
__________ = _________________________ = ___________ = ____
Angle BCE arctan (1/2) arctan (1/2) 1
