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 AE, BF. Shew that CECF2 = DE2 + DF2.

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

AD2 + 2.BD2 + 3. CD2.

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.

(b) 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 Fand 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.

BOOK II.

DEFINITIONS.

I.

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.

"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 BC be two straight lines;

and let BC' be divided into any parts BD, DE, EC, in the points D, E. Then the rectangle contained by the straight lines A and BC, 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.

Then the rectangle BH is equal to the rectangles BK, DL, EH.
And BH is contained by A and BC,

for it is contained by GB, BC, and GB is equal to 4:
and the rectangle BK is contained by A, BD,
for it is contained by GB, BD, of which GB is equal to A:
also DL is contained by A, DE,

because DK, that is, BG, (1. 34.) is equal to A;

and in like manner the rectangle EH 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 contained 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.

Upon AB 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 AF is the rectangle contained by BA, AC; for it is contained by DA, AC, of which DA is equal to 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.

PROPOSITION III. THEOREM.

If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square on the aforesaid part.

Let the straight line AB be divided into any two parts in the point C.
Then the rectangle AB, BC, shall be equal to the rectangle
AC, CB, together with the square on BC.

Upon BC describe the square CDEB, (1. 46.) and produce ED to F,
through 4 draw AF parallel to CD or BE, (1. 31.) meeting EF in F.
Then the rectangle AE is equal to the rectangles AD, CE.
And AE is the rectangle contained by AB, BC,

for it is contained by AB, BE, of which BE is equal to BC:
and AD is contained by AC, CB, for CD is equal to CB:
and CE is the square on BC:

therefore the rectangle AB, BC, is equal to the rectangle AC, CB, together with the square on BC.

If therefore a straight line be divided, &c.

PROPOSITION IV. THEOREM.

Q.E.D.

If a straight line be divided into any two parts, the square on the whole line is equal to the squares on the two parts, together with twice the rectangle contained by the parts.

Let the straight line AB be divided into any two parts in C.

Then the square on AB shall be equal to the squares on AC, and CB, together with twice the rectangle contained by AC, CB.

Upon AB describe the square ADEB, (1. 46.) join BD,

through C draw CGF parallel to AD or BE, (1. 31.) meeting BD in G and DE in F;

and through G draw HGK parallel to AB or DE, meeting AD in H, and BE in K;

Then, because CF is parallel to AD and BD falls upon them, therefore the exterior angle BGC is equal to the interior and opposite angle BDA; (1. 29.)

but the angle BÚA is equal to the angle DBA, (1. 5.)
because BA is equal to AD, being sides of a square;