14. If the quadrilateral ABCD is not cyclic, prove that the three rectangles AB.CD, BC.AD, AC.BD are proportional to the three sides of a triangle which has an angle equal to the sum of a pair of opposite angles of the quadrilateral.
15. Prove by using Theorem 11 that if perpendiculars be let fall on the sides and diagonals of a cyclic quadrilateral, from any point in the circumference of the circumscribed circle, the rectangle contained by the perpendiculars on the diagonals is equal to the rectangle contained by the perpendiculars on either pair of opposite sides.
16. If AB be the diameter of a semicircle, and PA, PB chords from any point P in the circumference, and if a perpendicular to AB from any point C meet PA, PB in D and E, and the semicircle in F, CF is a mean proportional between CD and CE.
PROP. XVIII.—Problem.
On a given right line (AB) to construct a rectilineal figure similar to a given one (CDEFG), and similarly placed as regards any side (CD) of the latter.
Def.—Similar figures are said to be similarly described upon given right lines, when these lines are homologous sides of the figures.
Sol.—Join CE, CF, and construct a triangle ABH on AB equiangular to CDE, and similarly placed as regards CD; that is, make, the angle ABH equal to CDE, and BAH equal to DCE. In like manner construct the triangle HAI equiangular to ECF, and similarly placed, and lastly, the triangle IAJ equiangular and similarly placed with FCG. Then ABHIJ is the figure required.
Dem.—From the construction it is evident that the figures are equiangular, and it is only required to prove that the sides about the equal angles are proportional. Now because the triangle ABH is equiangular to CDE, AB : BH :: CD : DE [iv.]; hence the sides about the equal angles B and D are proportional. Again, from the same triangles we have BH : HA :: DE : EC, and from the triangles IHA, FEC; HA : HI :: EC : EF; therefore (ex æquali) BH : HI :: DE : EF; that is, the sides about the equal angles BHI, DEF are proportional, and so in like manner are the sides about the other equal angles. Hence (Def. i.) the figures are similar.
Observation.—In the foregoing construction, the line AB is homologous to CD, and it is evident that we may take AB to be homologous to any other side of the given figure CDEFG. Again, in each case, it the figure ABHIJ be turned round the line AB until it falls on the other side, it will still be similar to the figure CDEFG. Hence on a given line AB there can be constructed two figures each similar to a given figure CDEFG, and having the given line AB homologous to any given side CD of the given figure.