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 A Midpoint Construction (Posted on 2015-05-15)

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.

 See The Solution Submitted by Bractals No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: Solution | Comment 2 of 3 |
(In reply to Solution by Harry)

Looks good except for one point -

When proving triangles NMC' and EBA are similar, I would use

/NC'M = /EAB   instead of  /NC'M = /BAM.

The reason is its false if point A lies between points M and E.

 Posted by Bractals on 2015-05-18 18:13:38

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