Fig. 46. Fig. 47. A c' c c' C c'. c c' e ; therefore, W = that is—When two the first ; and draw AC equal to the third, and making any angle with AB; join DC, and through B draw BE paral- portion that is, BA is a mean proportional LC; therefore—Each side of the original triangle is a e BD BA o * (BA)? . 101. The equation |BA. T BC’ gives BD = TECT * . DC A . CA)2 and the equation GA = BC, gives DC — o ad ding these two equations we obtain BD + DC = BC = BA)2 CA)2 (BC)* = (BA)2 + (AC)”. From this it is evident that if we refer to a common measure the three sides of a right-angled triangle, the second power of the number expressing the length of the hypothen use, will be equal to the sum of the second powers of the numbers which express the lengths of the other two sides. We can therefore find the hypothen use when we have the other sides. Suppose in this triangle that AB contains the common measure six times, and AC eight times; then (BC)* = 36 -–64 = 100; and taking the second root of I both sides of this equation, we have BC = (100)* = 10. So also if we know the hypothen use and one of the sides, we can find the other. Suppose we have the hypothenuse equal to 10, and the side AB equal to 6, to find the side AC. The equation (BC)2 = (BA)2 + (AC)”, by subtracting (BA)* from both sides, will give (AC)* = (BC)” – (BA)2 = 100 — 36 = 64; and taking the sec I ond root of each side, we obtain AC = (64)? = 8. 102. In the triangle ABC (fig. 48), bisect the angle B by the line BD; and through the point C, draw CE parallel to DB till it meets AB produced. The angle ABC, exterior to the triangle CBE, is equal to the sum of the two interior opposite angles BCE and BEC (37); but on account of the parallels DB and CE, the angle ABD (half of ABC) is equal to the angle E.; the other half must be equal to the angle BCE, consequently BCE and BEC will be equal ; therefore, the opposite sides BC and BE are equal (45). Now in the triangle AEC, DB is drawn parallel to the base, which Fig. 49. 3. # = ; (98); but BC is equal to BE, and may be substituted for it in the proportion; AB AD . BC T DC * secting an angle of a triangle, divides the opposite side into parts proportional to the adjacent sides. gives the proportion which gives, that is—A straight line bi 103. Let us now take an obtuse-angled triangle, as ABC (fig, 49) and draw perpendiculars from the verti- have BG T BE * proportion Multiplying in order, the two last proportions, gives if now we divide each and comparing this proportion with , We we perceive that they have the diculars drawn from the vertices to the opposide sides, would meet in the point C. We say therefore—In any triangle, the perpendiculars drawn from the vertices of the angles to the opposite sides, will meet in the same oint. p Remark. If the triangle be acute-angled, the point of meeting will be within the triangle ; if the triangle be obtuse-angled, the point of meeting will be without ; and in the right-angled triangle, the point of coincidence is the vertex of the right-angle. ' Of the Straight Line and Circle. 104. The circle is the plane surface embraced by the circular curve already described (7), called the circumference of the circle. A straight line which touches the circumference only in one point, is called a tangent, as AB (fig. 50). When it is said that the straight line touches the circumference, it is meant that its direction is such that if it were produced ever so far each way, it would have only one point in common with the circumference. This point is called the point of contact. - 105. If the straight line cuts the circumference in two points, it is called a secant. A straight line which has its two extremities in the circumference, as DE (fig. 51), is called a chord; it subtends both the arc and the angle at the centre measured by this arc. The portion of surface contained between the two radii DO, EO, and the arc DGE, is called a sector. The portion contained between the chord and the arc is called a segment of the circle. 106. If the two radii DO, EO, the chord DE and the arc DGE, be made to revolve together about the centre O, of the circle, the points D and E will be continually in the circumference ; and the arc DEG will always coincide with the part of the circumference then embraced by the two radii (7); if this were not the case, some parts of the circumference must be nearer to the centre than other parts; moreover, the straight line DE would be constantly the chord of this arc. Now as the angle DOE would not be changed by this revolution of the sector DOE, we say that this angle would have the same arc in every part of the circumference, and this arc would also have the same chord. We have, therefore, Fig. 50. Fig. 51. Fig. 52. these propositions:—(1). In the same circle or in equal |