(§ 66) .. BE = EF = CE = EG, and BF= CG. In Fig. 2. Produce AB and DC to meet at H. Since, by hyp., ≤ ABC > ≤ BCD, ≤ CBH < ▲ BCH. Lay off, on BH, FH = CH; and on DH, GH = BH; and draw line FG cutting BC at E. .. AFGH=▲ BCH. .. ZCBH =LFGH. Then, in ▲ BEF and CEG, ▲ EBF = ≤ CGE. (§ 63) (§ 66) (?) (§§ 86, 68) (§ 66) In Fig. 3. Produce BA and CD to meet at K. Since, by hyp., ABC > BCD, CK> BK. (?) Lay off, on KB produced, FK=CK; and on CK, GK=BK; and draw line FG cutting BC at E. since BF = FK – BK, and CG = CK – GK. .. Δ ΒΕΓ = Δ CEG. (?) = .. BE CE and EF = EG. (?) Then since, in either figure, BC + CG = BF + FG, and ▲ BEF=▲ CEG, quadrilateral AFGD is isoperimetric with, and to, quadrilateral ABCD. Calling the remainder of the given polygon P, it follows that the polygon composed of AFGD and P is isoperimetric with, and to, the polygon composed of ABCD and P; that is, the given polygon. Then the polygon composed of AFGD and P must be the maximum of polygons having the given perimeter and the given number of sides. Hence, the polygon composed of AFGD and P is equilateral. (§ 381) But this is impossible, since AF is > DG. In like manner, ABC cannot be <BCD. .. Z ABC = ▲ BCD. Note. The case of triangles was considered in § 380. Fig. 3 also provides for the case of triangles by supposing D and K to coincide with A. In the case of quadrilaterals, P=0. 383. Cor. Of isoperimetric polygons having the same number of sides, that which is regular is the maximum. PROP. V. THEOREM. 384. Of two isoperimetric regular polygons, that which has the greater number of sides has the greater area. Given ABC an equilateral ▲, and M an isoperimetric Proof. Let D be any point in side AB of ▲ ABC. Draw line DC; and construct isosceles ACDE isoperi metric with ▲ BCD, CD being its base. .. area CDE > area BCD. .. area ADEC > area ABC. But, since ADEC and M are isoperimetric, area Marea ADEC. .. area M> area ABC. (§ 379) (§ 381) In like manner, we may prove the area of a regular pentagon greater than that of an isoperimetric square; etc. 385. Cor. The area of a circle is greater than the area of any polygon having an equal perimeter. SYMMETRICAL FIGURES. DEFINITIONS. 386. Two points are said to be symmetrical with respect to a third, called the centre of symmetry, when the latter bisects the straight line which joins them. Thus, if O is the middle point of straight line AB, points A and B are symmetrical with respect to O A 0 as a centre. L B 387. Two points are said to be symmetrical with respect to a straight line, called the axis of symmetry, when the latter bisects at right angles the straight line which joins them. Thus, if line CD bisects line AB at right angles, points A and B are symmetrical with respect to CD as an axis. F 388. Two figures are said to be symmetrical with respect to a centre, or with respect to an axis, when to every point of one there corresponds a symmetrical point in the other. 389. Thus, if to every point of triangle ABC there corresponds a symmetrical point of triangle A'B'C', with respect to centre 0, triangle A'B'C' is symmetrical to triangle ABC with respect to centre 0. Again, if to every point of triangle ABC there corresponds a symmetrical point of triangle A'B'C", with respect to axis DE, triangle A'B'C' is symmetrical to triangle ABC with respect to axis DE. 390. A figure is said to be symmetrical with respect to a centre when Devery straight line drawn through the centre cuts the figure in two points which are symmetrical with respect to that centre. 391. A figure is said to be symmetrical with respect to an axis when it divides it into two figures which are symmetrical with respect to that axis. 392. Two straight lines which are symmetrical with respect to a centre are equal and parallel. B Given str. lines AB and A'B' symmetrical with respect to centre O. To Prove AB and A'B' equal and . Proof. Draw lines 44', BB', AB', and A'B. 393. If a figure is symmetrical with respect to two axes at right angles to each other, it is symmetrical with respect to their intersection as a centre. Given figure AE symmetrical with respect to axes XX' and YY, intersecting each other at rt. at O. To Prove AE symmetrical with respect to O as a centre. Proof. Let P be any point in the perimeter of AE. Draw line PQI XX', and line PR YY'. Produce PQ and PR to meet the perimeter of AE at P' and P", respectively, and draw lines QR, OP', and OP". Then since AE is symmetrical with respect to XX', PQ = P'Q. But PQ OR; whence, OR is equal and || to P'Q. = Therefore, OP'QR is a □. Whence, QR is equal and || to OP'. ($ 387) (?) (?) In like manner, we may prove OP"RQ a □; and therefore QR equal and || to OP". Then since both OP' and OP" are equal and || to QR, P'OP" is a str. line which is bisected at 0. That is, every str. line drawn through O is bisected at that point, and hence AE is symmetrical with respect to as a centre. (§ 390) |