+29 Just a quick note, this question has been posted already, but the answer wasn't fully explained in detail, so I'm reposting it in hopes that someone can explain one step about how to do the problem.
Let A=(0,0), B=(a,b), and C=(c,0) be points in the coordinate plane for some numbers a,b, and c such that AB=AC. Let D be the midpoint of BC, E be the foot of the perpendicular from D to CA, and F is the midpoint of DE.
(1) Find the coordinates of points E and F in terms of a,b, and c.
(2) Show that AF and BE are perpendicular.
The previous Solution did by CPhill at this link: https://web2.0calc.com/questions/geometry-proof_8
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If AB = AC
Then the coordinates of of C must be ( sqrt ( a^2 + b^2) , 0) = ( c , 0) (how did he get that ( sqrt ( a^2 + b^2) , 0) = ( c , 0)???)
But this implies that a^2 + b^2 = c^2
Which implies that a^2 - c^2 = -b^2
Which also means that (a - c) (a + c) = - b^2
D = [ ( a + c)/2 , b/2 ]
E = [( a + c) /2, 0]
F = [ (a+ c) / 2 , b/4 ]
Slope of AF = b/4 b
_________ = _______
(a + c)/2 2 (a + c)
Slope of BE = b b b 2b
_______________ = ___________ = _______ = ____
( a - (a+ c) / 2) a - a/2 - c/2 a/2 - c/2 a - c
Multiplying top.bottom of the last fraction by a + c we have that
2b ( a + c) 2b (a + c) 2 ( a + c) - 2 ( a + c)
____________ = ____________ = _________ = __________
(a - c) (a + c) - b^2 - b b
So slope of AF * slope of BE =
b - 2 (a + c)
_________ * _________ = - 1
2(a + c) b
Which means that AF and BE are perpendicular
