CHAPTER XIV. PROPORTION. SECTION I.-Theorems on Proportion. 117. If two circles intersect, the common chord produced bisects the common tangent. Let A B be the common chord, and let the points of contact of a tangent to the circles be T and P, and let A B produced meet T P in C. Because C is a point without the first circle, CT being a tangent and CA a secant, 118. The tangents drawn from a point to the same circumference are equal. Refer to Problem IX., page 126. The figure is symmetrical about AO, and therefore the two tangents AD, A D' are equal. 119. If two diagonals of a regular pentagon be drawn to cut one another, they will be divided in extreme and mean ratio, and the greater segment will be equal to the side of the pentagon. Let the diagonals AC, BE be drawn from the extremities of the side A B of the regular pentagon A B C D E, and intersect each other in the point H. Then BE and AC are cut in extreme and mean ratio in H, and the greater segment of each is equal to the side of the pentagon. Let the circle A B C D E be described about the pentagon. Because arc E A ... arc E A B = arc A Barc B C, arc A B C ; Therefore E B A C and ▲ BAC = LABE= = Z BEA, and consequently the triangles ABE, ABH are equiangular : therefore E B is to B A A B to B H. But BA is equal to E H, therefore E B is to E H = EH is to BH. But BE is greater than E H; therefore E H is greater than HB: Therefore BE has been cut in extreme and mean ratio in H. Similarly, it may be shewn that AC has also been cut in extreme and mean ratio in H, and that the greater segment of it, C H, is equal to the side of the pentagon. 120. If two equal chords of a circle cut one another either within or without a circle, the segments of the one between the point of intersection and the circumference shall be equal to the segments of the other, each to each. Refer to Example 68, page 66. Also let the lengths. of AE = a, A B = C D = m ; Then EB ma, and ED = m + c. Or else m = a + c, or a = m - C. Either result shows that the segments of the one are respectively equal to the segments of the other. 121. If tangents be drawn at the extremities of any two diameters of a circle, and produced to intersect one another, the straight lines joining the opposite points of intersection will both pass through the centre. The figure formed will be a parallelogram, the centre of which will be the centre of the circle. 122. If any chord of a circle be produced equally both ways and tangents to the circle be drawn on opposite sides of it from its extremities, the line joining the points of contact bisects the given chord. Produce CE to meet G D produced in F, and draw D H parallel to C F. Because FE = F G ; therefore ▲ F E G = / FGE. = But since HD is parallel to FE, DHG = FEG = DGH; therefore DH = DG. Again sqr. on CE = rect. CA C B, and sq. on DG rect. D B D A. |