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Angle Bisector in Touching Circles: What is this about?
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The applet attempts to illustrate a problem once counted among the Forty Exciting Problems by then college students, Frank D'Ippolito and Ravi Vakil ([Honsberger, pp. 71-74]):
Two circles are internally tangent at a point T. A chord AB of the outer circle touches the inner circle at the point P. Prove that TP always bisects  ATB.
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If AB is parallel to the common tangent at T, the configuration is symmetric with respect two the line of centers of the two circles, and the assertion is obvious. Assume then AB is inclined to the tangent line and crosses the latter in Q. Since QP and QT are both tangent to the inner circle, they are equal and ΔPQT is isosceles so that
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PTQ = TPQ.
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If we introduce x = ATP, y = PTB, z = BTQ, then
| (1) |
TPQ = y + z.
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On the other hand, TPQ is exterior to ΔATP. Thus,
| (2) |
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But TAB is subtended by arc BT enclosed by the tangent at T and chord BT. Therefore,
| (3) |
TAB = BTQ = z.
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Now we combine (1)-(3) to obtain
| (3) | y + z = x + z, |
and x = y, as required.
The problem admits a generalization wherein the circles do not have to be tangent and a further one wherein the tangent is replaced by a chord.
Nathan Bowler suggested another solution. The two circles are homothetic with center at T. Let point K correspond P under that homothety. On one hand, K lies on TP. On the other, the tangent at K is parallel to that at P which is but the chord AB. But tangent to an arc parallel to the subtending chord is bound to touch the arc at its midpoint. Thus, K is the midpoint of arc. Therefore, PT that passes through K bisects ATB.
References
- F. G.-M., Exercices de Géométrie, Éditions Jacques Gabay, sixiéme édition, 1991, pp. 277-278
- R. Honsberger, From Erdös To Kiev, MAA, 1996.
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