Saturday, 11 May 2013

geometry - Intersection of Altitudes in Hexagon




I've been trying to figure out how CEVA's Theorem can be implemented in solving this problem, but I'm coming up short and cannot make any progress with this problem. The problem states;



A convex hexagon ABCDEF satisfies |AB| = |BC|; |CD| =
|DE|; |EF| = |FA|. Prove that the lines containing the altitudes of the
triangles BCD, DEF, FAB starting respectively at the vertices C, E, A
intersect at a common point.



Any advice or guidance is much appreciated!


Answer



Draw the circles kB,kD,kF centered at the vertices B,D,F and radii BA=BC,DC=DE,FE=FA respectively. Then the altitude line hA of triangle ABF throguh vertex A, the altitude line hC of triangle BCD throguh vertex C and the altitude line hE of triangle DEF through vertex E are the radical axes of the three pairs of circles (kF,kB), (kB,kD) and (kD,kF) respectively. Therefore, by the radical axis theorem, the three radical axes hA,hC and hF intersect at a common point.

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