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A Midpoint Construction (Posted on 2015-05-15) Difficulty: 3 of 5

  
Let M be a point in the plane of ΔABC such that line AM
bisects ∠BAC. Let A', B', and C' be the perpendicular
projections of point M onto lines BC, CA, and AB respectively.
Let N be the intersection of lines A'M and B'C'.

Prove that line AN bisects side BC.
  

  Submitted by Bractals    
Rating: 4.0000 (1 votes)
Solution: (Hide)

Let D be the intersection of line AM and line B'C'.
Let E be the intersection of line AM and side BC.
Let F be the midpoint of side BC.

PRELIMINARY:

   (a)   ∠ABE = ∠C'MN    because BA'MC' is a cyclic
                          quadrilateral and the interior 
                          angle at a vertex is equal to 
                          the exterior angle at the 
                          opposite vertex.   

   (b)   ∠BAE = ∠MC'N    because the right triangles 
                          ADC' and C'DM are similar.  

   (c)   (a) & (b)  ==>  ΔC'MN ~ ΔABE by ASA.
                    ==>  |NC'||AB| = |MC'||AE|

   (d)   ∠ACE = ∠B'MN    because CB'MA' is a cyclic
                          quadrilateral and the interior 
                          angle at a vertex is equal to 
                          the exterior angle at the 
                          opposite vertex.   

   (e)   ∠CAE = ∠MB'N    because the right triangles
                          ADB' and B'DM are similar.  

   (f)   (d) & (e)  ==>  ΔB'MN ~ ΔACE by ASA.
                    ==>  |NB'||AC| = |MB'||AE|

                          |NC'|     |NB'|
   (g)   (c) & (f)  ==>  ------- = -------
                           |AC|      |AB|  

                          because |MB'| = |MC'|.

PROOF:       

Let PQ denote the vector from point P to point Q. 

Let s = +1 if C' lies on ray AB
      = -1 if not.

   AN = AC' + C'N

                |C'N|
      = AC' + -------- C'B' 
               |C'B'|

                |NC'|
      = AC' + --------*( AB' - AC')
               |B'C'|

                |NC'|            |NC'|
      = ( 1 - -------- ) AC' + -------- AB'
               |B'C'|           |B'C'|

          |NB'|          |NC'|
      = -------- AC' + -------- AB'
         |B'C'|         |B'C'|

         |NB'|*s*|AC'|        |NC'|*s*|AB'|
      = --------------- AB + --------------- AC
           |B'C'||AB|           |B'C'||AC|

            Because |AC'| = |AB'|  and the
            PRELIMINARY

         |NB'|*s*|AB'|        
      = ---------------*( AB + AC )
           |B'C'||AB|      

         2*|NB'|*s*|AB'|        
      = -----------------*( AB + AC )/2
            |B'C'||AB|    

         2*|NB'|*s*|AB'|        
      = -----------------*AF
            |B'C'||AB|  

Therefore, since line AN coincides with vector AN,
line AN bisects side BC.

QED    

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re(2): SolutionHarry2015-05-19 07:59:17
re: SolutionBractals2015-05-18 18:13:38
SolutionSolutionHarry2015-05-18 13:22:12
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