NOTATION: PQ denotes the vector from point P to point Q.
Since A is a right angle, AC·AD = 0.
Since E lies on line CD, AE = xAC + (1-x)AD for some real x.
Since lines AE and CD are perpendicular,
AE·CD = [ xAC + (1-x)AD ]·[ AD - AC ]
= xAC·AD - x|AC|2 + (1-x)|AD|2 - (1-x)AC·AD
= (1-x)|AD|2 - x|AC|2 = 0
Since F is the midpoint of CE,
AF = ( AC + AE )/2
= [ (1+x)AC + (1-x)AD ]/2
Therefore,
AF·BE = AF·( AE - AB ) = AF·( AE - 2AD )
= [ (1+x)AC + (1-x)AD ]·[ xAC - (1+x)AD ]/2
= [ x(1+x)|AC|2 - (1+x)2AC·AD + x(1-x)AC·AD - (1-x)(1+x)|AD|2 ]/2
= -(1+x)[ (1-x)|AD|2 - x|AC|2 ]/2 = 0
Therefore, BE and AF are perpendicular.
Sorry that the bold does not show up in pre blocks.
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