equiangular; and therefore TA is to TB as AC is to BD (VI. 4, V. 16), that is, as AK is to BM.
Therefore the triangles TAK and TBM are similar (YI. 7); therefore the angle TAK is equal to the angle TBM; and therefore AK is parallel to BM. Similarly AL is parallel to BN. And because AK is parallel to BM and AG parallel to BD the angle CAK is equal to the angle DBM; and therefore the angle CLK is equal to the angle DNM (III. 20); and therefore CL is parallel to DN. Similarly CK is parallel to DM.
Now TM is to TD as TD is to TN (III. 37, VI. 16); and TM is to TD as TK is to TC (VI. 4); therefore TK is to TC as TD is to TN; and therefore the rectangle TK, TN is equal to the rectangle TC, TD, Similarly the rectangle TL, TM is equal to the rectangle TC, TD.
If each of the given circles is without the other we may suppose the straight line which touches both circles to meet AB at T between A and B, and the above results will all hold, provided we interchange the letters K and L; so that the five letters are now to be in the following order, L, K, T, M, N
The point T is called a centre of similitude of the two circles.
