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Intersection of Bisectors (Posted on 2012-12-01) Difficulty: 3 of 5
 
Let Γ be a circle with center O. Let AC be a chord of  Γ (not containing O).
Let D be a point on chord AC (not A or C). The line through D and perpendicular to AC intersects Γ in points B1 and B2. Let H1 and H2 be the orthocenters of triangles
AB1C and AB2C respectively.

Define the point of intersection of the bisectors of angles H1B1O and H2B2O and
prove your result.
 

See The Solution Submitted by Bractals    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Proof. Comment 2 of 2 |
(In reply to From sketchpad. No proof. by Jer)

  • The point of intersection is on the circle.
  • It is independent of point D.
  • It is the point where the perpendicular bisector of segment AC intersects the circle (on the same side of the center.)

Proof

Since the segments B1D and B2D are both altitudes of triangles and are also collinear and since the orthocenters are on these altitudes, H1 and H2 are both on this same line.
[So the solution is independent of D]

One of angles AB1C and AB2C is acute and the other is obtuse.  So lets make AB1C be the acute one.
Now H1 is between B1 and D, but B2 is between H2 and D.

Consider triangle B1OB2.   It is isosceles so let angles OB1B2 = OB2B1 = a and angle B1OB2 = 180-2a

Call the point of interest X
Now angles OB1B2 and OB2H2 are bisected by rays through X forming triangle B1XB2.
Angle XB1B2 = a/2
Angle OB2H2 = 180-a so angle XB2H2 = 90-a/2 so angle XB2B1 = 90-a
This makes angle B1XB2 = 90-a

The significance is since B1OB2 is a central angle and angle B1XB2 is half of this, X must lie on the circle.

Finally triangle XOB2 is isosceles so angle OXB2= 90-a/2
subtract this from angle B1XB2 to get angle OXB1 = a/2
angles OXB1 and XB1B2 are alternate interior angles.  Since they are equal, segments OX and B1B2 are parallel.
So we have my final assertion: OX is perpendicular to AC.
  Posted by Jer on 2012-12-03 13:56:48
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