# Three Points Casey's Theorem

Circle (S) is tangent to the circumcircle of points A, B, C. AT_{a}, BT_{b}, CT_{c} are tangent to (S). Assume that M - the point of tangency of the two circles - is in the arc AB. Prove that

BC·AT_{a} + AC·BT_{b} = AB·CT_{c}.

What if applet does not run? |

### References

- D. O. Shklyarsky, N. N. Chentsov, I. M. Yaglom,
*Selected Problems and Theorems of Elementary Mathematics*, v 2, Moscow, 1952, #123(a).

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Copyright © 1996-2018 Alexander BogomolnyCircle (S) is tangent to the circumcircle of points A, B, C. AT_{a}, BT_{b}, CT_{c} are tangent to (S). Assume that M - the point of tangency of the two circles - is in the arc AB. Prove that

BC·AT_{a} + AC·BT_{b} = AB·CT_{c}.

What if applet does not run? |

### 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' = (AT_{a})²,

BM·BB' = (BT_{b})²,

CM·CC' = (CT_{c})²,

such that

AM² / (AT_{a})² = AM / AA',

BM² / (BT_{b})² = BM / BB',

CM² / (CT_{c})² = CM / CC',

implying

AM / AT_{a} = BM / BT_{b} = CM / CT_{c} = √ρ / (r ± ρ),

which simply shows that the required identity

BC·AT_{a} + AC·BT_{b} = AB·CT_{c}.

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

BC·AM + AC·BM = AB·CM.

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Copyright © 1996-2018 Alexander Bogomolny