25. If, in Fig. 437, PM = k . PN + 1, where k and I are constants, and PMO and PNO are any constant angles, then the locus of P is a straight line. 26. Construct a right-angled triangle such that one side shall be a mean proportional between the hypotenuse and the other side. 27. A, B, C, D are collinear points and on AC and BD two similar triangles APC and BQD are described, so that the corresponding sides AP and BQ, and again CP and DQ, are parallel. If QP and DA intersect in 0, prove that OA: OB = OC: OD. 28. The line joining any point P on a given circle to a fixed point Q is divided in a given ratio at R. Prove that the locus of R is a circle. 29. ABCD and EFGH are two equiangular quadrilaterals, and AB : EF = CD: GH. Prove that the quadrilaterals are similar. Is there any exception to this? = = 30. ABCD is a quadrilateral, and P, Q, R, S are points on AB, BC, CD, DA such that AP: AB = CQ: CB: CR: CD AS: AD. Prove that PORS is a parallelogram. 31.* If the vertex A of a triangle ABC be joined to any point D in the base BC, then the circum-radii of triangles ADB and ADC are proportional to AB and AC. 32.* ABCD is a parallelogram, and a line is drawn from A cutting BD in P, CD in Q, and BC produced in R. Prove that PQ: PR=PD2 : PB2, and that AP is a mean proportional to PQ and PR. 33.* If a triangle of given species has one angle fixed, and another angle moves along a given straight line (or a given circle) then the third angle moves along a fixed straight line (or a fixed circle). 34. Given two straight lines AB and CD, find a point P such that PAB and PCD are similar triangles. SIM. GEO. FF CHAPTER XXI. PROPORTION THEOREMS ON AREAS 306. THEOREM [Ar. 6].—Triangles (or parallelograms) of equal altitude are proportional to their bases; and triangles (or parallelograms) on equal bases are proportional to their altitudes. (i) In As ABC, DEF (Fig. 442) :— Given that altitude AH = altitude DK, Required to prove that A ABC: A DÉF = BC: EF. Cons. Assuming that BC and EF are commensurable, there is some length P which is contained an exact number of times in each of them. Suppose BC and EF divided respectively into r and s parts each equal to P. In each triangle join the points of section of the base to the vertex. A D B L M N H C FIG. 442. Proof.-As ABL, ALM, AMN, Ar. 5. Δ ABL; on equal bases and of equal altitudes. Hence they are all of equal area. Thus ▲ ABC = r. ▲ ABL, and ▲ DEF = 8. Required to prove that A ABC: A DEF = altitude AH: altitude DK. Cons. Let HM KN BC (= EF); join AM, DN. = = : As AMH, DNK have equal altitudes MH, NK, ..they are proportional to their bases AH, DK. Proved. AAMH: A DNK = AH : DK; i.e. hence A ABC: A DEF = AH : DK. (iii) The proofs for parallelograms are obvious. 307. Alternative Proof of Theorem Ar. 6 :— Let b1, b2 be the bases of the two triangles; h1, h2 their respective altitudes; and A1, A2 their respective areas. Then by Theorem Ar.3, Cor. 1, A1 = § b1 h1, A2 = { b2h2. Δι b1h bi ხა 01 h1 Δε h2 308. On Algebraical Operations in Geometry. It is often convenient in geometrical argument to allow the symbol AB to denote either the line AB or the measure of the line AB in terms of some unit of length, and similarly to allow the symbol ▲ ABC to denote either the triangle or its measure in terms of some unit of area. The value of this practice lies in the fact that any mathematical operation (such as multiplication or the extraction of a root) may be performed with the measures, as these are abstract numbers. This practice does not lead to confusion of ideas or to incorrect argument. It may be used in either of the following cases : CASE I. In equations involving only ratios. ΔΑΒΗ and A CDK have the same value, whether they refer to the ratios of these lengths and areas, or whether they refer to the ratios of the measures of these lengths and areas. CASE II. In equations involving mensuration formulæ. For theorem Ar.1 is correctly expressed by the equation rect. BADC AB × AD, = provided that the measures of these quantities are expressed in corresponding units of length and area; also all other mensuration formulæ in areas are derived from this formula. EXERCISES CXVII (Riders). 1. If straight lines AO, BO, CO be drawn from the vertices of a triangle ABC to any point 0 within the triangle, and if AO produced cuts BC in L, prove that 2. ABCD is a quadrilateral, and its diagonals AC and BD intersect in 0. Show that A AOB: A BOC = 3. ABC is a triangle. Find a point P in ABC such that the triangles PAB, PBC, and PCA are all equal. 4. A quadrilateral is divided into four triangles by its diagonals; show that if two of these triangles be equal the remaining two are either equal or similar. 5. In Figs. 417, 418, 419, join BE, CF; then prove T. 11 and T.11c by means of Ar. 6. 309. THEOREM [Ar.7.].-If two triangles (or two parallelograms) have one angle of the one equal to one angle of the other, their areas are in the ratio of the products of the sides containing the equal angles. LB = LE. (i) In Fig. 444, Required to prove that AABC: A DEF AB. BC: DE. EF. = Cons. Let A DEF be placed so that the arms of the angle E lie along the arms of the equal angle B: and let GBH represent A DEF in its new position. Join GC. Да B H CE F FIG. 444. As GBC, GBH on bases BC, BH, have the same (ii) The corresponding theorem for parallelograms follows at once from the fact that either diagonal bisects a parallelogram. NOTE. In the above argument we have used the notation explained and justified in § 308. |