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.
(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 nonCartesian 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 20121227 14:05:53 