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Pagina 122 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. D c A' D' Hyp. In triangles ABC and A'B'C', ZA = ZA'. To prove AABC = ABxAC. A A'B'C' A'B'xA'C' Proof. Draw the altitudes BD and B'D'.
Pagina 79 - If a straight line touch a circle, and from the point of contact a chord be drawn, the angles which this chord makes with the tangent are equal to the angles in the alternate segments.
Pagina 125 - ... figures are to one another in the duplicate ratio of their homologous sides.
Pagina 43 - DE : but equal triangles on the same base and on the same side of it, are between the same parallels ; (i.
Pagina 86 - BFE : (i. def. 10.) therefore, in the two triangles, EAF, EBF, there are two angles in the one equal to two angles in the other, each to each ; and the side EF, which is opposite to one of the equal angles in each, is common to both ; therefore the other sides are equal ; (i.
Pagina 113 - A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment as the greater segment is to the less.
Pagina 52 - If a straight line be bisected, and produced to any point, the square of the whole line thus produced, and the square of the part of it produced, are together double of the square of half the line bisected, and of the square of the line made up of the half and the part produced.
Pagina 56 - Iff a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line which is made up of the whole and that part.
Pagina 74 - Upon the same straight line, and upon the same side of it, there cannot be two similar segments of circles, not coinciding with one another. If it be possible. let the two similar segments of circles, viz. ACB' ADB be upon the same side of the same straight line AB, not coinciding with one another.