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 Perpendicular (Posted on 2012-12-26)
Let ABC be a triangle and D the foot of the altitude from A. Let E and F lie on a line passing through D such that AE is perpendicular to BE, AF is perpendicular to CF, and E and F are different from D. Let M and N be the midpoints of the segments BC and EF, respectively. Prove that AN is perpendicular to NM.

 No Solution Yet Submitted by Danish Ahmed Khan Rating: 3.3333 (3 votes)

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 Possible approach | Comment 2 of 5 |

Note that as well as being 'a point on BC', D is, much more significantly, a vertex of right triangles ACD, ABD and ADM. Now if E lies on (the) line passing through D such that AE is perpendicular to BE, it follows that E is a point on the circle whose diameter is AB, (and D is on that circle too, because it is a vertex of right triangle ABD) and  if F lies on (the same) line passing through D such that AF is perpendicular to CF, it follows that F is a point on the circle whose diameter is AC (and D is on that circle too, because it is a vertex of right triangle ACD).

Last, if N is the modpoint of EF, then N is a point on the circle whose diameter is AM, (and D is a point on that circle too, because it is a vertex of right triangle ADM).

But If so, then angle ANM is a right angle, as was to be shown.

 Posted by broll on 2012-12-26 23:57:56

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