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# Analytic Geometry question

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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

Nov 26, 2020