to a learner, who is unacquainted with the properties of the circle. The following example, however, will illustrate this part of the subject sufficiently for our purpose at present. Find a point in a given straight line, such that the sum of its distances from two fixed points on the same side of the line is a minimum, that is, less than the sum of the distances of any other point in the line from the fixed points. Taking the diagram of the last example, suppose CD to be the given line, and A, B the given points. Now if A and A' be symmetrical with respect to CD, we know that every point in CD is equally distant from A and A'. (See Note VIII. p. 103.) Hence the sum of the distances of any point in CD from A and B is equal to the sum of the distances of that point from A' and B. But the sum of the distances of a point in CD from A' and B is the least possible when it lies in the straight line joining A' and B. Hence the point P, determined as in the last example, is the point required. NOTE. Propositions IX., X., XI., XII. of Book I. give good examples of symmetrical constructions. NOTE XI. Euclid's Prop. V. of Book I. The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced, the angles upon the other side of the base shall be equal. Let ABC be an isosceles A, having AB= AC. Produce AB, AC to D and E. Then must ABC= L ACB, and LDBC= L ECB. .: FA=GA, and AC=AB, and ▲ FAC= 4 GAB, :. FC=GB, and AFC= LAGB, and ▲ ACF= L ABG. : BF=CG, and FC=GB, and BFC= CGB, .. LFBC= 4 GCB, and ▲ BCF= 4 CBG. of which the parts ▲ BCF and 4 CBG are equal; .. remaining ACB=remaining ABC. Also it has been proved that FBC= ↳ GCB, that is, LDBC= L ECB. I. 4. Ax. 3. Q. E. D. NOTE XII. Euclid's Prop. VI. of Book I. If two angles of a triangle be equal to one another, the sides also which subtend the equal angles shall be equal to one another. For if not, AB is either greater or less than AC. From AB cut off BD=AC. Then in As DBC, ACB, · DB=AC, and BC is common, and ▲ DBC= L ACB, .. ADBC= ▲ ACB ; that is, the less = the greater; which is absurd. .. AB is not greater than AC. I. 4. Similarly it may be shewn that AB is not less than AC; .. AB AC. NOTE XIII. Euclid's Prop. VII. of Book I. Q. E. D. Upon the same base and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and their sides which are terminated in the other extremity of the base equal also. If it be possible, on the same base AB, and on the same side of it, let there be two as ACB, ADB, such that AC=AD, and also BC=BD. Join CD. First, when the vertex of each of the As is outside the that is, BDC is both equal to and greater than BCD; which is absurd. Secondly, when the vertex D of one of the As falls within the other ▲ (fig. 2); Then Produce AC and AD to E and F. ::: AC=AD, .. LECD= L FDC. But ECD is greater than 4 BCD; Again, that is, .. FDC is greater than ▲ BCD ; much more is 4 BDC greater than 4 BCD. I. 5. BDC is both equal to and greater than 4 BCD; which is absurd. Lastly, when the vertex D of one of the As falls on a side BC of the other, it is plain that BC and BD cannot be equal. Q. E. D. NOTE XIV. Euclid's Prop. VIII. of Book I. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal, the angle which is contained by the two sides of the one must be equal to the angle contained by the two sides of the other. DA B E Let the sides of the ▲s ABC, DEF be equal, each to each, that is, AB=DE, AC=DF and BC= EF. Apply the Then Then must L BAC LEDF. ABC to the ▲ DEF, so that pt. B is on pt. E, and BC on EF. ::: BC=EF, .. C will coincide with F, and BC will coincide with EF. Then AB and AC must coincide with DE and DF. For if AB and AC have a different position, as GE, GF, then upon the same base and upon the same side of it there can be two As which have their sides which are terminated in one extremity of the base equal, and their sides which are terminated in the other extremity of the base also equal: which is impossible. .. since base BC coincides with base EF, AB must coincide with DE, and AC with DF; .. BAC coincides with and is equal to EDF. I. 7. Q.E.D. |