Consider two directly similar triangles $△A_1B_1C_1$ and $△A_2B_2C_2$ with $$B_1C_1:A_1C_1:A_1B_1=B_2C_2:A_2C_2:A_2B_2=a:b:c. $$ Then $a·A_1A_2, b·B_1B_2$ and $c·C_1C_2$ form the sides of a triangle. The triangle is degenerate if and only if the similitude center of $△A_1B_1C_1$ and $△A_2B_2C_2$ lies on the circumcircle of these triangles.
This theorem extends Ptolemy's theorem and the Ptolemy inequality.
Note on Mr Tweedie's Theorem in Geometry, Peter Pinkerton Proceedings of the Edinburgh Mathematical Society, Volume 22 - February 1903 Published online by Cambridge University Press: 20 January 2009 note-on-mr-tweedies-theorem-in-geometry.pdf(58.95 KB, Downloads: 44)
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Let ABC, A′B′C′ (Fig. 4) be two triangles equiangular in the same sense. Let BC, B′C′ meet in X. Describe circles round BXB′, CXC′ to meet again in O. Then it is easy to see that the triangles BOC, COA, AOB are equiangular in the same sense to the triangles B′OC′, C′OA′, A′OB′ respectively. Hence the triangles AOA′, BOB′, COC′ are similar; $\therefore \frac { AA ^ { \prime } } { AO } = \frac { BB ^ { \prime } } { BO } = \frac { CC } { CO }$ $\therefore a . {AA}', b . {B B}', c \cdot {C C}'$ are proportional to $a . {A O}, b . {B O}, c . {C O}$ where $a, b, c$ are the sides of the triangle $A B C$.
From ${O}$ draw OP, OQ, OR perpendicular to BC, CA, AB respectively. Then ${Q R}={A O} \sin {A} \propto a . {A O}$, ${RP}={BO} \sin {B} \propto b . {BO}$, ${PQ}={CO} \sin {C} \propto c . {CO}$; $\therefore a . {A A}', b . {BB}', c . {C C}'$, being proportional to $a . {A O}, b . {BO}, c . {CO}$, are proportional to $Q R, R P, P Q$. But PQR is a triangle, unless O is on the circumcircle of ABC when PQR is the Simson line of O. $\therefore QR + RP>PQ$, with two similar inequalities, except that one of the inequalities becomes an equality if O is on the circumcircle of ABC. $\therefore a. AA' + b. BB'>c.CC'$, with two similar inequalities; one of the inequalities becoming an equality when O lies on the circumcircle of ABC. Similarly in the case of an equality O lies also on the circumcircle of A'B'C. For the case of equilateral triangles $a = b = c$ ; $\therefore AA' + BB'>CC'$, with two similar inequalities ; one of the three inequalities becoming an equality when O lies on the circumcircles of ABC and A'B'C. It is obvious that the theorem reduces to Ptolemy's Theorem or its converse.
