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and AB produced in G; join AF, CG; then will the triangles ABF, CFG be equal to one another.
123. ABCD is a parallelogram, E the point of intersection of its diagonals, and K any point in AD. If KB, KC be joined, shew that the figure BKEC is one-fourth of the parallelogram.
124. Let ABCD be a parallelogram, and ( any point within it, through O draw lines parallel to the sides of ABCD, and join 0A, OC; prove that the difference of the parallelograms DO, BO is twice the triangle O AC.
125. The diagonals AC, BD of a parallelogram intersect in O, and P is a point within the triangle A OB; prove that the difference of the triangles APB, CPD is equal to the sum of the triangles A PC, BPD.
126. If K be the common angular point of the parallelograms about the diameter AC (fig. Euc. I. 43.) and BD be the other diameter, the difference of these parallelograms is equal to twice the triangle BKD.
127. The perimeter of a square is less than that of any other parallelogram of equal area.
128. Shew that of all equiangular parallelograms of equal perimeters, that which is equilateral is the greatest.
129. Prove that the perimeter of an isosceles triangle is greater than that of an equal right-angled parallelogram of the same altitude.
VIII. 130. If a quadrilateral figure is bisected by one diagonal, the second diagonal is bisected by the first.
131. If two opposite angles of a quadrilateral figure are equal, . shew that the angles between opposite sides produced are equal.
132. Prove that the sides of any four-sided rectilinear figure are together greater than the two diagonals.
133. The sum of the diagonals of a trapezium is less than the sum of any four lines which can be drawn to the four angles, from any point within the figure, except their intersection.
134. The longest side of a given quadrilateral is opposite to the shortest; shew that the angles adjacent to the shortest side are together greater than the sum of the angles adjacent to the longest side.
135. Give any two points in the opposite sides of a trapezium, inscribe in it a parallelogram having two of its angles at these points.
136. Shew that in every quadrilateral plane figure, two parallelograms can be described upon two opposite sides as diagonals, such that the other two diagonals shall be in the same straight line and equal.
137. Describe a quadrilateral figure whose sides shall be equal to four given straight lines. What limitation is necessary ?
138. If the sides of a quadrilateral figure be bisected and the points of bisection joined, the included figure is a parallelogram, and equal in area to half the original figure.
* 139. A trapezium is such, that the perpendiculars let fall on a diagonal from the opposite angles are equal. Divide the trapezium into four equal triangles, by straight lines drawn to the angles from a point within it.
140. If two opposite sides of a trapezium be parallel to one another, the straight line joining their bisections, bisects the trapezium.
141. If of the four triangles into which the diagonals divide a trapezium, any two opposite ones are equal, the trapezium has two of its opposite sides parallel.
142. If two sides of a quadrilateral are parallel but not equal, and the other two sides are equal but not parallel, the opposite angles of the quadrilateral are together equal to two right angles: and conversely.
143. If two sides of a quadrilateral be parallel, and the line joining the middle points of the diagonals be produced to meet the other sides; the line so produced will be equal to half the sum of the parallel sides, and the line between the points of bisection equal to half their difference.
144. To bisect a trapezium, (1) by a line drawn from one of its angular points: (2) by a line drawn from a given point in one side.
145. To divide a square into four equal portions by lines drawn from any point in one of its sides.
146. It is impossible to divide a quadrilateral figure (except it be a parallelogram) into equal triangles by lines drawn from a point within it to its four corners.
IX. 147. If the greater of the acute angles of a right-angled triangle, be double the other, the square on the greater side is three times the square on the other.
148. Upon a given straight line construct a right-angled triangle such that the square on the other side may be equal to seven times the square on the given line..
149. If from the vertex of a plane triangle, a perpendicular fall upon the base or the base produced, the difference of the squares on the sides is equal to the difference of the squares on the segments of the base.
150. If from the middle point of one of the sides of a right-angled triangle, a perpendicular be drawn to the hypotenuse, the difference of the squares on the segments into which it is divided, is equal to the square on the other side.
151. If a straight line be drawn from one of the acute angles of a right-angled triangle, bisecting the opposite side, the square upon that line is less than the square upon the hypotenuse by three times the square upon half the line bisected.
152. If the sum of the squares on the three sides of a triangle be equal to eight times the square on the line drawn from the vertex to the point of bisection of the base, then the vertical angle is a right angle.
153. If a line be drawn parallel to the hypotenuse of a rightangled triangle, and each of the acute angles be joined with the points where this line intersects the sides respectively opposite to them, the squares on the joining lines are together equal to the squares on the hypotenuse and on the line drawn parallel to it.
154. Let ACB, ADB be two right-angled triangles having a common hypotenuse AB, join CD, and on CD produced both ways draw perpendiculars A E, BF. Shew that CE + CF?=DE? + DF?.
155. If perpendiculars AD, BE, CF drawn from the angles on the opposite sides of a triangle intersect in G, the squares on AB, BC, and CA, are together three times the squares on AG, BG, and CG.
156. If ABC be a triangle of which the angle A is a right angle; and BE, CF be drawn bisecting the opposite sides respectively: shew that four times the sum of the squares on BE and CF is equal to five times the square on BC.
157. If ABC be an isosceles triangle, and CD be drawn perpendicular to AB; the sum of the squares on the three sides is equal to
AD?+ 2.BD? + 3. CD!. 158. The sum of the squares described upon the sides of a rhombus is equal to the squares described on its diameters.
159. A point is taken within a square, and straight lines drawn from it to the angular points of the square, and perpendicular to the sides; the squares on the first are double the sum of the squares on the last. Shew that these sums are least when the point is in the center of the square.
160. In the figure Euc. I. 47,
(a) Shew that the diagonals FA, AK of the squares on AB, AC, lie in the same straight line.
(6) If DF, EK be joined, the sum of the angles at the bases of the triangles BFD, CEK is equal to one right angle.
(c) If BG and CH be joined, those lines will be parallel.
(d) If perpendiculars be let fall from F and K on BC produced, the parts produced will be equal; and the perpendiculars together will be equal to BC.
(e) Join GH, KE, FD, and prove that each of the triangles so formed, equals the given triangle ABC.
(f) The sum of the squares on GH, KE, and FD will be equal to six times the square on the hypotenuse.
(g) The difference of the squares on AB, AC, is equal to the difference of the squares on AD, AE.
161. The area of any two parallelograms described on the two sides of a triangle, is equal to that of a parallelogram on the base, whose side is equal and parallel to the line drawn from the vertex of the triangle, to the intersection of the two sides of the former parallelograms produced to meet.
162. If one angle of a triangle be a right angle, and another equal to two-thirds of a right angle, prove from the First Book of Euclid, that the equilateral triangle described on the hypotenuse, is equal to the sum of the equilateral triangles described upon the sides which contain the right angle.
EVERY right-angled parallelogram is called a rectangle, and is said to be contained by any two of the straight lines which contain one of the right angles.
II. In every parallelogram, any of the parallelograms about a diameter together with the two complements, is called a gnomon.
B G “ Thus the parallelogram HG together with the complements AF, FC, is the gnomon, which is more briefly expressed by the letters AGK, or EHC, which are at the opposite angles of the parallelograms which make the gnomon.'
PROPOSITION I. THEOREM. 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 A and B C be two straight lines; and let B C be divided into any parts BD, DE, EC, in the points D, E.
Then the rectangle contained by the straight lines A and B C, shall be equal to the rectangle contained by A and BD, together with that contained by A and D.E, and that contained by A and EC.
From the point B, draw BF at right angles to BC, (1. 11.)
and make BG equal to A ; (1. 3.)
through G draw GH parallel to BC, (1. 31.) and through D, E, C, draw DK, EL, CH parallel to BG, meeting
GH in K, L, H.
And BH is contained by A and BC,
and the rectangle BK is contained by A, BD,
also ĎL is contained by A, DE, because DK, that is, BG, (1. 31.) is equal to A; and in like manner the rectangle ÈH is contained by A, EC: therefore the rectangle contained by A, BC, is equal to the several
rectangles contained by A, BD, and by A, DE, and by A, EC. Wherefore, if there be two straight lines, &c. Q. E. D.
PROPOSITION II. THEOREM.
If a straight line be divided into any two parts, the rectangles containeil by the whole and each of the parts, are together equal to the square on the whole line.
Let the straight line AB be divided into any two parts in the point C.
Then the rectangle contained by AB, BC, together with that contained by AB, AC, shall be equal to the square on AB.
DF E Upon A B describe the square ADEB, (1. 46.) and through C draw CF parallel to AD or BE, (1. 31.) meeting DE in F.
Then AE is equal to the rectangles AF, CE.
And AE is the square on AB;
and CE is contained by AB, BC,
for BE is equal to AB: therefore the rectangle contained by AB, AC, together with the rectangle AB, BC is equal to the square on AB.
If therefore a straight line, &c. Q. E. D.