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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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 re: Possible approach (Spoiler) | Comment 4 of 5 |
(In reply to Possible approach by broll)

Broll. I’m not sure how you get from ‘N being the midpoint of EF’ to ‘N is a point on the circle with diameter AM’ ?

This is the shortest non-Cartesian way I can find:

/ AEB = / ADB = 90     =>                  AEDB is a cyclic quadrilateral.

/ AFC = / ADC = 90     =>                  AFCD is a cyclic quadrilateral.

/ BAE            =  / CDF      (external angle  of cyclic quad)
=  / CAF      (angles in same segment)

Adding / EAC to each of these gives:             / BAC = / EAF
Also:                                                                                     / ABC = / AEF                  (external angle of cyclic quad.)

So triangles ABC and AEF are similar and, since M and N are mid points of corresponding sides, triangles ABM and AEN are also similar, giving   / AMB = / ANE.

Now, since both these angles are subtended by AD, it follows that ANMD is a cyclic quad.
and therefore that:   / ANM        =  / ADB     (external angle of cyclic quad.)
=  90 degrees.

 Posted by Harry on 2012-12-27 14:05:53

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