Exercises.
Def. i.—Homologous points in the planes of two similar figures are such, that lines drawn from them to the angular points of the two figures are proportional to the homologous sides of the two figures.
1. If two figures be similar, to each point in the plane of one there will be a corresponding point in the plane of the other.
Dem.—Let ABCD, A′B′C′D′ be the two figures, P a point in the plane of ABCD. Join AP, BP, and construct a triangle A′P′B′ on A′B′, similar to APB; then it is easy to see that lines from P′ to the angular points of A′B′C′D′ are proportional to the lines from P to the angular points of ABCD.
2. If two figures be directly similar, and in the same plane, there is in the plane a special point which, regarded as belonging to either figure, is its own homologous point with respect to the other. For, let AB, A′B′ be two homologous sides of the figures, C their point of intersection. Through the two triads of points A, A′, C; B, B′, C describe two circles intersecting again in the point O: O will be the point required. For it is evident that the triangles OAB, OA′B′ are similar and that either may be turned round the point O, so that the two bases, AB, A′B′, will be parallel.
Def. ii.—The point O is called the centre of similitude of the figures. It is also called their double point.
3. Two regular polygons of n sides each have n centres of similitude.
4. If any number of similar triangles have their corresponding vertices lying on three given lines, they have a common centre of similitude.