Homothety and Spiral Similarity

Since \(AD:DB=2:3\), we have \(AD:AB=2:5\). Because \(DE \parallel BC\), triangles \(ADE\) and \(ABC\) are similar. The similarity ratio is \(\frac{AD}{AB}=\frac{2}{5}\). Therefore \(\frac{DE}{BC}=\frac{2}{5}\), so \(DE:BC=2:5\).

Since \(AB \parallel CD\), we have \(\angle OAB=\angle OCD\) and \(\angle OBA=\angle ODC\). Hence \(\triangle OAB \sim \triangle OCD\). Thus \(\frac{OA}{OC}=\frac{OB}{OD}=\frac{AB}{CD}\). Points \(A,C,O\) and \(B,D,O\) lie on the corresponding lines, so the homothety with centre \(O\) and ratio \(\frac{OC}{OA}\) sends \(A\) to \(C\), \(B\) to \(D\), and \(AB\) to \(CD\).

For external tangency, points \(O_1,T,O_2\) are collinear, with \(TO_1=3\) and \(TO_2=7\). The homothety with centre \(T\) and ratio \(-\frac{7}{3}\) sends centre \(O_1\) to \(O_2\) and the circle of radius \(3\) to the circle of radius \(7\). Therefore \(T\) is the centre of such a homothety, and \(TO_1:TO_2=3:7\).

In triangles \(PAC\) and \(PBD\), the sides adjacent to the angles at \(P\) are proportional: \(\frac{PA}{PB}=\frac{PC}{PD}\). The included angles are equal: \(\angle APC=\angle BPD\). Therefore \(\triangle PAC \sim \triangle PBD\). Hence \(P\) may be viewed as the centre of a spiral similarity sending \(AC\) to \(BD\).

Triangles \(OAB\) and \(OCD\) share the angle at \(O\). The adjacent sides are proportional: \(\frac{OA}{OC}=\frac{OB}{OD}\). Therefore \(\triangle OAB \sim \triangle OCD\). From similarity, \(\angle OAB=\angle OCD\), so lines \(AB\) and \(CD\) are parallel.

Angles \(\angle AOD\) and \(\angle COB\) are vertical. Since \(AD \parallel BC\), we get \(\angle ADO=\angle CBO\). Hence \(\triangle AOD \sim \triangle COB\). From similarity, \(\frac{AO}{OC}=\frac{DO}{OB}=\frac{AD}{BC}\), in particular \(\frac{AO}{OC}=\frac{DO}{OB}\).

A homothety sends a circle to a circle and preserves the central symmetry of the figure. Therefore the centre of the first circle is sent to the centre of the second circle: \(O_1 \mapsto O_2\). Under a homothety, a point, its image, and the centre of homothety are collinear. Hence \(H,O_1,O_2\) are collinear.

Let \(H\) be the external centre of homothety, with the smaller circle closer to \(H\). Then \(\frac{HO_1}{HO_2}=\frac{4}{10}=\frac{2}{5}\). Since the centre is external, points \(H,O_1,O_2\) are collinear and \(HO_2=HO_1+18\). Let \(HO_1=x\). Then \(\frac{x}{x+18}=\frac{2}{5}\). Hence \(5x=2x+36\), so \(x=12\). Therefore \(HO_1=12\).

In the similarity \(\triangle PAC \sim \triangle PBD\), the corresponding vertices are \(P \leftrightarrow P\), \(A \leftrightarrow B\), \(C \leftrightarrow D\). Therefore corresponding sides are proportional: \(\frac{AC}{BD}=\frac{PA}{PB}=\frac{PC}{PD}\). Corresponding angles are equal, so \(\angle ACP=\angle BDP\).

Rotate the plane about \(P\) by angle \(\angle APB\), sending ray \(PA\) to ray \(PB\). Then apply a dilation with ratio \(\frac{PB}{PA}\). Point \(A\) goes to \(B\). Since \(\angle APB=\angle CPD\), the same rotation sends ray \(PC\) to ray \(PD\). From \(\frac{PA}{PB}=\frac{PC}{PD}\), we get \(\frac{PB}{PA}=\frac{PD}{PC}\), so the same dilation sends \(C\) to \(D\). Therefore \(P\) is the centre of the spiral similarity.

The homothety with centre \(A\) and ratio \(\frac{1}{2}\) sends \(B\) to \(D\), \(C\) to \(E\), and segment \(BC\) to \(DE\). A midpoint is sent to the midpoint of the image. Therefore the midpoint \(M\) of \(BC\) is sent to the midpoint \(N\) of \(DE\). A point, its image, and the centre of homothety are collinear, so \(A,M,N\) are collinear.

Since points \(A,B,C,M\) lie on one circle, \(\angle BMC=\angle BAC\). Since points \(A,D,E,M\) lie on one circle, \(\angle DME=\angle DAE\). But \(AB\) and \(AD\) are one line, and \(AC\) and \(AE\) are another line. Hence \(\angle BAC=\angle DAE\). Therefore \(\angle BMC=\angle DME\).

Since \(AD \parallel BC\), triangles \(AOD\) and \(COB\) are similar. Therefore \(\frac{AD}{BC}=\frac{AO}{OC}=\frac{3}{2}\). Hence \(\frac{21}{BC}=\frac{3}{2}\), so \(BC=14\).

Triangles \(AOD\) and \(COB\) have vertical angles at \(O\). By the condition, \(\frac{AO}{OC}=\frac{DO}{OB}\), so the sides adjacent to these angles are proportional. Therefore \(\triangle AOD \sim \triangle COB\). Hence \(\angle DAO=\angle BCO\). These are alternate interior angles for lines \(AD\) and \(BC\), so \(AD \parallel BC\).

From \(AB \parallel CD\), we get \(\triangle OAB \sim \triangle OCD\), so \(O\) is the centre of the homothety sending \(AB\) to \(CD\), with ratio \(\frac{OC}{OA}\). The image of \(X\) under this homothety lies both on line \(OX\) and on line \(CD\), so it is point \(Y\). Therefore \(\frac{OY}{OX}=\frac{OC}{OA}\), hence \(\frac{OX}{OY}=\frac{OA}{OC}\).

Sides \(PA,PC\) and \(PB,PD\) are proportional, and the included angles are equal. Therefore \(\triangle PAC \sim \triangle PBD\). Hence corresponding angles are equal: \(\angle PAC=\angle PBD\). Also, from similarity, \(\frac{AC}{BD}=\frac{PA}{PB}\).

A rotation about \(P\) by angle \(\angle APB\) sends ray \(PA\) to ray \(PB\). By the condition, the same angle sends ray \(PC\) to ray \(PD\). Then a dilation with centre \(P\) and ratio \(2\) sends \(A\) to \(B\), because \(PB=2PA\), and sends \(C\) to \(D\), because \(PD=2PC\). Thus one rotation followed by one dilation sends \(AC\) to \(BD\), so \(P\) is the centre of the spiral similarity.

Let \(H\) be the internal centre of homothety. Then \(H\) lies on segment \(O_1O_2\) and \(\frac{HO_1}{HO_2}=\frac{6}{9}=\frac{2}{3}\). Let \(HO_1=x\), so \(HO_2=20-x\). We get \(\frac{x}{20-x}=\frac{2}{3}\). Hence \(3x=40-2x\), so \(5x=40\), \(x=8\). Therefore \(HO_1=8\).

The angle between \(MB\) and \(MC\) is \(\angle BMC\), and the angle between \(MD\) and \(ME\) is \(\angle DME\). Since \(A,B,C,M\) lie on one circle, \(\angle BMC=\angle BAC\). Since \(A,D,E,M\) lie on one circle, \(\angle DME=\angle DAE\). Lines \(BAD\) and \(CAE\) determine the same angle at \(A\), so \(\angle BAC=\angle DAE\). Therefore the required angles are equal.

From the similarity \(\triangle PAC \sim \triangle PBD\), we get \(\angle CAP=\angle DBP\). Since \(AC \parallel PB\), angle \(\angle CAP\) equals the angle between \(PB\) and \(PA\). Therefore the angle between \(DB\) and \(PB\) equals the angle between \(PA\) and \(PB\). Hence lines \(BD\) and \(PA\) are parallel.

Let the external tangents touch the first circle at \(A_1,B_1\) and the second at \(A_2,B_2\), with \(A_1,A_2,H\) on one tangent and \(B_1,B_2,H\) on the other. Radii to points of tangency are perpendicular to the tangents, so triangles \(HO_1A_1\) and \(HO_2A_2\) are right triangles and share the angle at \(H\). Hence they are similar. Thus \(\frac{HO_1}{HO_2}=\frac{O_1A_1}{O_2A_2}\). The same homothety with centre \(H\) sends the radius of the first circle to the corresponding radius of the second. Therefore it sends the first circle to the second and \(O_1\) to \(O_2\). Hence \(H,O_1,O_2\) are collinear.

From the cyclicity of \(A,B,C,M\), we have \(\angle MBC=\angle MAC\). From the cyclicity of \(A,D,E,M\), we have \(\angle MDE=\angle MAE\). Since \(A,C,E\) are collinear, \(\angle MAC=\angle MAE\), so \(\angle MBC=\angle MDE\). Also, from the previous configuration, \(\angle BMC=\angle DME\). Therefore \(\triangle MBC \sim \triangle MDE\). By the condition \(MB=MD\), the similarity ratio is \(1\), so \(MC=ME\).

Since \(D_iE_i \parallel BC\), segment \(D_iE_i\) is the image of \(BC\) under a homothety with centre \(A\) and some ratio \(k_i\). Under this homothety, the midpoint \(M\) of \(BC\) is sent to the midpoint \(M_i\) of \(D_iE_i\). Therefore points \(A,M,M_i\) are collinear for \(i=1,2\). Hence \(A,M_1,M_2,M\) are collinear.

We have \(\frac{PA}{PB}=\frac{6}{9}=\frac{2}{3}\) and \(\frac{PC}{PD}=\frac{10}{15}=\frac{2}{3}\). By the condition, \(\angle APC=\angle BPD\). Hence \(\triangle PAC \sim \triangle PBD\) by two proportional sides and the included angle. Therefore \(P\) is the centre of the spiral similarity sending \(AC\) to \(BD\). From similarity, \(\frac{AC}{BD}=\frac{PA}{PB}=\frac{2}{3}\), so \(AC:BD=2:3\).