# Three Points Casey's Theorem

Circle (S) is tangent to the circumcircle of points A, B, C. ATa, BTb, CTc are tangent to (S). Assume that M - the point of tangency of the two circles - is in the arc AB. Prove that

BC·ATa + AC·BTb = AB·CTc.

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Solution

### References

1. D. O. Shklyarsky, N. N. Chentsov, I. M. Yaglom, Selected Problems and Theorems of Elementary Mathematics, v 2, Moscow, 1952, #123(a). Circle (S) is tangent to the circumcircle of points A, B, C. ATa, BTb, CTc are tangent to (S). Assume that M - the point of tangency of the two circles - is in the arc AB. Prove that

BC·ATa + AC·BTb = AB·CTc.

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### Solution

The problem is just a particular instance of Casey's theorem, wherein three of the four circles degenerate into points. Still, I'll give a proof independent of Casey's theorem.

Point M serves the center of similitude of the two circles. If A', B', C' are the second points of intersection of AM, BM, CM, with (S) then

A'M/AM = B'M/BM = C'M/CM = ρ/r

where r and ρ are the radii of the circumcircle (ABC) and that of (S), respectively. It follows that

AA'/AM = BB'/BM = CC'/CM = (r ± ρ)/ρ.

(The sign is chosen depending on whether the circumcircle (ABC) is outside or inside (S).)

On the other hand, by the Power of a Point Theorem,

AM·AA' = (ATa)²,
BM·BB' = (BTb)²,
CM·CC' = (CTc)²,

such that

AM² / (ATa)² = AM / AA',
BM² / (BTb)² = BM / BB',
CM² / (CTc)² = CM / CC',

implying

AM / ATa = BM / BTb = CM / CTc = ρ / (r ± ρ),

which simply shows that the required identity

BC·ATa + AC·BTb = AB·CTc.

is a direct consequence of that of Ptolemy's applied to the quadrilateral ABCM:

BC·AM + AC·BM = AB·CM. ### Ptolemy's Theorem 