Angle Bisectors in Ellipse II

Let A and B be two points on an ellipse with foci E and F. The tangents to the ellipse at A and B meet in S. Prove that ∠ASE = ∠BSF. In words, at the point of intersection, the two tangents to an ellipse are equally inclined to the lines joining that point with the foci of the ellipse.

22 January 2016, Created with GeoGebra

Proof

Conic Sections > Ellipse

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Let A and B be two points on an ellipse with foci E and F. The tangents to the ellipse at A and B meet in S. Prove that ∠ASE = ∠BSF.

Proof

To see why this is so, draw an ellipse through S confocal to the given one. This reminds of the configuration which showed the ellipse as an envelope of a family of straight lines.

angle bisector in ellipse

So we now have two confocal ellipses: ellipse1 (the original one) and ellipse2 (passing through S.) The configuration admits the following interpretation. Draw a tangent to ellipse1 at A till it hits ellipse2 in S. Reflect it at S and continue to the next intersection with ellipse2, and so on. All so constructed lines will touch an ellipse confocal with ellipse2 which is bound to be ellipse1 since the latter, by the construction, is already tangent to the first line AS. It follows that the second line in the chain is necessarily BS, implying that AS and BS are equally inclined to the tangent to ellipse2 at S. But so are ES and FS, and we are done.

Conic Sections > Ellipse

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  • Property of Angle Bisectors
  • Property of Angle Bisectors II
  • A Property of Angle Bisectors III
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  • Angle Bisectors On Circumcircle
  • Angle Bisectors in a Quadrilateral - Cyclic and Otherwise
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  • Triangle From Angle Bisectors
  • Property of Internal Angle Bisector - Hubert Shutrick's PWW
  • Angle Bisectors Cross Circumcircle
  • For Equality Choose Angle Bisector
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