base BC is bisected in D: let any line EDG be drawn through D, meeting AC the greater side in G and AB produced in E, and forming a triangle AEG having the same vertical angle A. Draw BH parallel to AC, and the triangles BDH, GDC are equal. Euc. 1. 26.

111. Let two triangles be constructed on the same base with equal perimeters, of which one is isosceles. Through the vertex of that which is not isosceles draw a line parallel to the base, and intersecting the perpendicular drawn from the vertex of the isosceles triangle upon the common base. Join this point of intersection and the extremities of the base.

112. (1) DF bisects the triangle ABC (fig. Prop. 6, p. 73.) On each side of the point F in the line BC, take FG, FH, each equal to one-third of BF, the lines DG, DH shall trisect the triangle. Or,

Let ABC be any triangle, D the given point in BC. Trisect BC in E, F. Join AD, and draw EG, FH parallel to AD. Join DG, DH; these lines trisect the triangle. Draw AE, AF and the proof is manifest.

(2) Let ABC be any triangle; trisect the base BC in D, E, and join AD, AE. From D, E, draw DP, EP parallel to AB, AC and meeting in P. Join AP, BP, CP; these three lines trisect the triangle.

(3) Let P be the given point within the triangle ABC. Trisect the base BC in D, E. From the vertex A draw AD, AE, AP. Join PD, draw AG parallel to PD and join PG. Then BGPA is one-third of the triangle. The problem may be solved by trisecting either of the other two sides and making a similar construction.

113. The base may be divided into nine equal parts, and lines may be drawn from the vertex to the points of division. Or, the sides of the

triangle may be trisected, and the points of trisection joined.

114. It is proved, Euc. 1. 34, that each of the diagonals of a parallelogram bisects the figure, and it may be shewn that they also bisect each other. It is hence manifest that any straight line, whatever may be its position, which bisects a parallelogram, must pass through the intersection of the diagonals.

115. See the remark on the preceding problem 114.

116. Trisect the side AB in E, F, and draw EG, FH parallel to AD or BC, meeting DC in G and H. If the given point P be in EF, the two lines drawn from P through the bisections of EG and FH will trisect the parallelogram. If P be in FB, a line from P through the bisection of FH will cut off one-third of the parallelogram, and the remaining trapezium is to be bisected by a line from P, one of its angles. If P coincide with E or F, the solution is obvious.

117. Construct a right-angled parallelogram by Euc. 1. 44, equal to the given quadrilateral figure, and from one of the angles, draw a line to meet the opposite side and equal to the base of the rectangle, and a line from the adjacent angle parallel to this line will complete the rhombus. 118. Bisect BC in D, and through the vertex A, draw AE parallel to BC, with center D and radius equal to half the sum of AB, AC, describe a circle cutting AE in E.

119. Produce one side of the square till it becomes equal to the diagonal, the line drawn from the extremity of this produced side and parallel to the adjacent side of the square, and meeting the diagonal producd, determines the point required.

120. Let fall upon the diagonal perpendiculars from the opposite angles of the parallelogram. These perpendiculars are equal, and each pair of triangles is situated on different sides of the same base and has equal altitudes. If the point be not on the diagonal, draw through the given point, a line parallel to a side of the parallelogram.

121. One case is included in Theo. 120. The other case, when the point is in the diagonal produced, is obvious from the same principle. 122. The triangles DCF, ABF may be proved to be equal to half of the parallelogram by Euc. 1. 41.

123. Apply Euc. 1. 41, 38.

124. If a line be drawn parallel to AD through the point of intersection of the diagonal, and the line drawn through O parallel to AB; then by Euc. 1. 43, 41, the truth of the theorem is manifest.

125. It may be remarked that parallelograms are divided into pairs of equal triangles by the diagonals, and therefore by taking the triangle ABD equal to the triangle ABC, the property may be easily shewn.

126. The triangle ABD is one half of the parallelogram ABCD, Euc. 1. 34. And the triangle DKC is one half of the parallelogram CDHG, Euc. I. 41, also for the same reason the triangle AKB is one half of the parallelogram AHGB: therefore the two triangles DKC, AKB are together one half of the whole parallelogram ABCD. Hence the two triangles DKC, AKB are equal to the triangle ABD: take from these equals the equal parts which are common, therefore the triangle CKF is equal to the triangles AHK, KBD: wherefore also taking AHK from these equals, then the difference of the triangles CKF, AHK is equal to the triangle KBC: and the doubles of these are equal, or the difference of the parallelograms CFKG, AHKE is equal to twice the triangle KBD.

127. First prove that the perimeter of a square is less than the perimeter of an equal rectangle: next, that the perimeter of the rectangle is less than the perimeter of any other equal parallelogram.

128. This may be proved by shewing that the area of the isosceles triangle is greater than the area of any other triangle which has the same vertical angle, and the sum of the sides containing that angle is equal to the sum of the equal sides of the isosceles triangle.

129. Let ABC be an isosceles triangle (fig. Euc. 1. 42), AE perpendicular to the base BC, and AECG the equivalent rectangle. Then AC is greater than AE, &c.

130. Let the diagonal AC bisect the quadrilateral figure ABCD. Bisect AC in E, join BE, ED, and prove BE, ED in the same straight line and equal to one another.

131. Apply Euc. 1. 15.

132. Apply Euc. 1. 20.

133. This may be shewn by Euc. 1. 20.

134. Let AB be the longest and CD the shortest side of the rectangular figure. Produce AD, BC to meet in E. Then by Euc. 1. 32.

135. Let ABCD be the quadrilateral figure, and E, F, two points in the opposite sides AB, CD, join EF and bisect it in G; and through G draw a straight line HGK terminated by the sides AD, BC; and bisected in the point G. Then EF, HK are the diagonals of the required parallelogram.

136. After constructing the figure, the proof offers no difficulty.

137. If any line be assumed as a diagonal, if the four given lines taken two and two be always greater than this diagonal, a four-sided figure may be constructed having the assumed line as one of its diagonals: and it may be shewn that when the quadrilateral is possible, the sum of every three given sides is greater than the fourth.

138. Draw the two diagonals, then four triangles are formed, two on one side of each diagonal. Then two of the lines drawn through the points of bisection of two sides may be proved parallel to one diagonal, and two

R

parallel to the other diagonal, in the same way as Theo. 97, supra. The other property is manifest from the relation of the areas of the triangles made by the lines drawn through the bisections of the sides.

139. It is sufficient to suggest, that triangles on equal bases, and of equal altitudes, are equal.

140. Let the side AB be parallel to CD, and let AB be bisected in E and CD in F, and let EF be drawn. Join AF, BF, then Euc. 1. 38.

141. Let BCED be a trapezium of which DC, BE are the diagonals intersecting each other in G. If the triangle DBG be equal to the triangle EGC, the side DE may be proved parallel to the side BC, by Euc. 1. 39. 142. Let ABCD be the quadrilateral figure having the sides AB, CD, parallel to one another, and AD, BC equal. Through B draw BE parallel to AD, then ABED is a parallelogram.

143. Let ABCD be the quadrilateral having the side AB parallel to CD. Let E, F be the points of bisection of the diagonals BD, AC, and join EF and produce it to meet the sides AD, BC in G and H. Through H draw LHK parallel to DA meeting DC in L and AB produced in K. Then BK is half the difference of DC and AB.

144. (1) Reduce the trapezium ABCD to a triangle BAE by Prob. 106, supra, and bisect the triangle BAE by a line AF from the vertex. If F fall without BC, through F draw FG parallel to AC or DE, and join AG.

Or thus. Draw the diagonals AC, BD: bisect BD in E, and join AE, EC. Draw FEG parallel to AC the other diagonal, meeting AD in F, and DC in G. AG being joined, bisects the trapezium.

(2) Let E be the given point in the side AD. Join EB. Bisect the quadrilateral EBCD by EF. Make the triangle EFG equal to the triangle EAB, on the same side of EF as the triangle EAB. Bisect the triangle EFG by EH. EH bisects the figure.

145. If a straight line be drawn from the given point through the intersection of the diagonals and meeting the opposite side of the square; the problem is then reduced to the bisection of a trapezium by a line drawn from one of its angles.

146. If the four sides of the figure be of different lengths, the truth of the theorem may be shewn. If, however, two adjacent sides of the figure be equal to one another, as also the other two, the lines drawn from the angles to the bisection of the longer diagonal, will be found to divide the trapezium into four triangles which are equal in area to one another. Euc. 1. 38.

147. Apply Euc. 1. 47, observing that the shortest side is one half of the longest.

148. Find by Euc. 1. 47, a line the square on which shall be seven times the square on the given line. Then the triangle which has these two lines containing the right angle shall be the triangle required.

149. Apply Euc. 1. 47.

150. Let the base BC be bisected in D, and DE be drawn perpendicular to the hypotenuse AC. Join AD: then Euc. 1. 47.

151. Construct the figure, and the truth is obvious from Euc. 1. 47. 152. See Theo. 32, p. 76, and apply Euc. 1. 47.

153. Draw the lines required and apply Euc. 1. 47.

154. Apply Euc. 1. 47.

155. Apply Euc. 1. 47.

156

Apply Euc. 1. 47, observing that the square on any line is four

times the square on half the line.

157. Apply Euc. 1. 47, to express the squares on the three sides in terms of the squares on the perpendiculars and on the segments of AB. 158. By Euc. 1. 47. bearing in mind that the square described on any line is four times the square described upon half the line.

159. The former part is at once manifest by Euc. 1. 47. Let the diagonals of the square be drawn, and the given point be supposed to coincide with the intersection of the diagonals, the minimum is obvious. Find its value in terms of the side.

160. (a) This is obvious from Euc. 1. 13.

(b) Apply Euc. 1. 32, 29.

(c) Apply Euc. 1. 5, 29.

(d) Let AL meet the base BC in P, and let the perpendiculars from F, K meet BC produced in M and N respectively; then the triangles APB, FMB may be proved to be equal in all respects, as also APC, CKN.

(e) Let fall DQ perpendicular on FB produced. Then the triangle DQB may be proved equal to each of the triangles ABC, DBF; whence the triangle DBF is equal to the triangle ABC.

Perhaps however the better method is to prove at once that the triangles ABC, FBD are equal, by shewing that they have two sides equal in each triangle, and the included angles, one the supplement of the other.

(f) If DQ be drawn perpendicular on FB produced, FQ may be proved to be bisected in the point B, and DQ equal to AC. Then the square on FD is found by the right-angled triangle FQD. Similarly, the square on KE is found, and the sum of the squares on FD, EK, GH will be found to be six times the square on the hypotenuse.

(g) Through A draw PAQ parallel to BC and meeting DB, EC produced in P, Q. Then by the right angled triangles.

161. Let any parallelograms be described on any two sides AB, AC of a triangle ABC, and the sides parallel to AB, AC be produced to meet in a point P. Join PA. Then on either side of the base BC, let a parallelogram be described having two sides equal and parallel to AP. Produce AP and it will divide the parallelogram on BC into two parts respectively equal to the parallelograms on the sides. Euc. 1. 35, 36.

162. Let the equilateral triangles ABD, BCE, CAF be described on AB, BC, CA, the sides of the triangle ABC having the right angle at A. Join DC, AK: then the triangles DBC, ABE are equal. Next draw DG perpendicular to AB and join CG: then the triangles BDG, DAG, DGC are equal to one another. Also draw AH, EK perpendicular to BC; the triangles EKH, EKA are equal. Whence may be shewn that the triangle ABD is equal to the triangle BHE, and in a similar way may be shewn that CAF is equal to CHE.

The restriction is unnecessary: it only brings AD, AE into the same

line.

GEOMETRICAL EXERCISES ON BOOK II.

6. See the figure Euc. 11. 5.

HINTS, &c.

7. This Problem is equivalent to the following: construct an isosceles right-angled triangle, having given one of the sides which contains the right angle.

8. Construct the square on AB, and the property is obvious.

9. The sum of the squares on the two parts of any line is least when the two parts are equal.

10. A line may be found the square on which is double the square on the given line. The problem is then reduced to :-having given the hypotenuse and the sum of the sides of a right-angled triangle, construct the triangle.

11. This follows from Euc. 11. 5, Cor.

12. This problem is, in other words, Given the sum of two lines and the sum of their squares, to find the lines. Let AB be the given straight line, at B draw BC at right angles to AB, bisect the angle ABC by BD. On AB take AE equal to the side of the given square, and with center A and radius AE describe a circle cutting BD in D, from D draw DF perpendicular to AB, the line AB is divided in F as was required.

13. Let AB be the given line. Produce AB to C making BC equal to three times the square on AB. From BA cut off BD equal to BC; then D is the point of section such that the squares on AB and BD are double of the square on AD. 14. In the fig. Euc. 11. 7. Join BF, and draw FL perpendicular on GD. Half the rectangle DB, BG, may be proved equal to the rectangle AB, BC.

Or, join KA, CD, KD, CK. Then CK is perpendicular to BD. And the triangles CBD, KBD are each equal to the triangle ABK. Hence, twice the triangle ABK is equal to the figure CBKD; but twice the triangle ABK is equal to the rectangle AB, BC; and the figure CBKD is equal to half the rectangle DB, CK, the diagonals of the squares on AB, BC. Wherefore, &c.

15. The difference between the two unequal parts may be shewn to be equal to twice the line between the points of section.

16. This proposition is only another form of stating Euc. II. 7.

17. In the figure, Theo. 7, p. 74, draw PQ, PR, PS perpendiculars on AB, AD, AC respectively: then since the triangle PAC is equal to the two triangles PAB, PAD, it follows that the rectangle contained by PS, AC, is equal to the sum of the rectangles PQ, AB, and PR, AD. When is the rectangle PS, AC equal to the difference of the other two rectangles?

18. Through E draw EG parallel to AB, and through F, draw FHK parallel to BC and cutting EG in H. Then the area of the rectangle is made up of the areas of four triangles; whence it may be readily shewn that twice the area of the triangle AFE, and the figure AGHK is equal to the area ABCD.

19. Apply Euc. II. 11.

20. The vertical angles at L may be proved to be equal, and each of them a right angle.

21. Apply Euc. 11. 4, 11. 1. 47.

22. Produce FG, DB to meet in L, and draw the other diagonal LHC, which passes through H, because the complements AG, BK are equal. Then LH may be shewn to be equal to Ff, and to Dd.

23. The common intersection of the three lines divides each into two parts, one of which is double of the other, and this point is the vertex of three triangles which have lines drawn from it to the bisection of the bases. Apply Euc. II. 12, 13.

24. Apply Theorem 3, p. 114, and Euc. 1. 47.

25. This will be found to be that particular case of Euc. I. 12, in which the distance of the obtuse angle from the foot of the perpendicular,