Other editions - View all
AB-AC AB=AC AB=BC AC produced altitude angle is equal Apply Ex base BC BC meets bisector Bookwork CD meet concurrent Construct a rectangle Construct a triangle convex polygon cut off equal diagonals intersect distances drawn 1 BC drawn parallel equal to twice equilateral triangle EUCLID Find the locus four points given points given square given straight line half the line hypotenuse internally at H isosceles triangle joining the mid-points line be divided line be drawn lines is equal medians meet AC mid-point of AB mid-points of BC obtuse angle opposite side parallel straight lines parallelogram perimeter perpendiculars drawn point in AB point in BC point of intersection points in order PQR is drawn produced if necessary quadrilateral ABCD rectangle contained rhombus right-angled triangle segments show that AB2 square on half Standard Theorem straight line PQ supplementary angles trapezium twice the rectangle twice the square whole line
Page 180 - If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.
Page 181 - IF a straight line be bisected, and produced to any point: the rectangle contained by the whole line thus produced, and the part of it produced, together with the square of half the line bisected, is equal to the square of the straight line which is made up of the half and the part produced.
Page 181 - If 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.
Page 180 - If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line. Let...
Page 169 - In every triangle, the square on the side subtending either of the acute angles, is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the...
Page 181 - To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square on the other part.
Page 181 - In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle, by twice the rectangle contained by the side...
Page 168 - In any obtuse triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides, increased by twice the product of one of these sides and the projection of the other side upon it.
Page 171 - AB into two parts, so that the rectangle contained by the whole line and one of the parts, shall be equal to the square on the other part.