EXERCISES ON BOOK ii.

NOTE-These Exercises are all Theorems to be proved; and depend mainly on the principles of Book ii.

1. The square on an altitude of an equilateral triangle is equal to threefourths the square on a side.

2. If any point within a rectangle is joined to its corners, the sum of the squares on the joins to a pair of opposite corners is equal to the sum of the squares on the other pair of joins.

NOTE―Join the point to the intersection of the diagonals, and use ii. Addenda (11).

3. If AXB, AYB are right-angled triangles on same side of same hypotenuse AB; and AP, BQ are perpendiculars on XY produced, then

XP2 + XQ2 = YP2 + YQ2.

4. If the hypotenuse AB of a right-angled triangle ABC is trisected in X, Y; then

NOTE-Use ii. Addenda (11), and Cor. ii. 4 (a).

5. If ABC is an isosceles triangle, and XY is parallel to the base BC; then, if BY is joined,

6. Any rectangle is equal to half the rectangle under the diagonals of the squares on two of its adjacent sides.

7. If from any point perpendiculars are dropped on all the sides of any rectilineal figure, the sum of the squares on the alternate segments are equal. 8. If ABC is any triangle, X a point in BC such that

AB2 + BX2

=

AC2 + CX2,

and M the mid point of AX; then BM

= CM.

9. If AB is the diameter of a circle; X, Y points in AB equidistant from the centre; P any point on the circumference; then

PX2 + PY2 = AX2 + AY2

=

BX2 + BY2.

NOTE-Use ii. Addenda (11), and ii. 10.

10. If ABCD is a parallelogram such that BA

=

BD; then

BD2+2 BC2 AC2.

=

II. If a line is divided in medial section, the sum of the squares on the whole line and on the lesser part is equal to three times the square on the

other part.

12. If the square on an altitude of a triangle is equal to the rectangle under the segments into which it divides the base, then the vertical angle is right.

13. Conversely to the last exercise-If a triangle is right-angled, then the square on the altitude from the corner of the right angle is equal to the rectangle under the segments into which it divides the hypotenuse.

14. ABC is an isosceles triangle, whose vertex is A: if CX is the perpendicular on AB; and XP the perpendicular on BC; then

15. In any triangle ABC, if BP, CQ are perpendiculars on AC, AB (produced if necessary) then

16. If the extremities of any chord of a circle are joined to any point in the diameter parallel to the chord; then the sum of the squares on the joins is equal to the sum of the squares on the segments of the diameter. 17. If ABCD is a square, and X a point in AC such that AX

figure ABXD = 2 AX 2.

=

AC; then

18. If BX, CY are squares on sides BA, CA of any triangle ABC, then

BC2 + XY2 = 2 (AB2 + AC2).

NOTE-Draw AN 1 to XY; and let NA meet BC in M: draw BP, CQ to NAM.

19. If squares are described on three sides of any triangle, and their corners joined; then the sum of the squares on the hexagon thus formed is equal to four times the sum of the squares on the sides of the triangle.

NOTE-Use preceding Exercise.

20. The sum of the squares on the diagonals of any quadrilateral is equal to twice the squares on the joins of the mid points of opposite sides.

NOTE-Use i. Addenda (18), and Cor. ii. 4, (a).

21. If two sides of a quadrilateral are parallel, then the sum of the squares on its diagonals is equal to the sum of the squares on its non-parallel sides together with twice the rectangle under its parallel sides.

=

22. If A, B, C, D are four collinear points; X the mid point of AB, Y the mid point of CD, and M the mid point of XY; and if P is any point; then PA2 + PB2 + PC2 + PD2 MA2 + MB2 + MC2 + MD2 + 4 PM2. NOTE-Use ii. 9, and ii. Addenda (11). Or deduce from property of mean centre on p. 113.

23. ABCD is any quadrilateral; the mid points of its diagonals are joined, and M is the mid point of this join; if P is any point; then

PA2 + PB + PC2 + PD2 = MA2 + MB2 + MC2 + MD2 + 4 PM2. 24. In the figure of ii. 11; if DX meets BE in Y, and (when produced) meets BF in Z; then DZ is perpendicular to BF and to AY.

25. If X, Y, Z are the feet of any concurrent perpendiculars on the sides, and D, E, F the mid points of the sides, respectively opposite corners A, B, C of a triangle; then, of the rectangles under BC, XD, under CA, YE, and under AB, ZF, the greatest is equal to the sum of the other two.

NOTE-Use ii. 13, and ii. Addenda (11).

26. If ABCD is a rectangle, X any point in BC, and Y any point in CD; then ABCD 2 A AXY + BX. DY.

=

27. If two opposite sides of a quadrilateral are bisected, then the sum of the squares on the other sides together with the squares on the diagonals is equal to the sum of the squares on the bisected sides together with four times the square on the join of their mid points.

28. In any triangle ABC, X, Y, Z are the feet of the altitudes, and O the orthocentre; then

AB2+ BC2 + CA2 = 2 (AX. AO + BY. BO + CZ.CO). NOTE-Use ii. 12.

29. If a point is taken within a triangle at which its three sides subtend equal angles; then the sum of the squares on the sides of the triangle is equal to twice the sum of the squares on the joins of its corners to that point, together with the sum of the rectangles under these joins taken two and two.

NOTE--Draw a from one of the corners on one of the joins.

30. In any quadrilateral the sum of the squares on the four lines from the middle of the join of the mid points of a pair of opposite sides to the corners of the quadrilateral, is equal to the sum of the squares on the joins of the mid points of opposite sides and the join of the mid points of the diagonals.

BOOK iii.

Def. A straight line terminated at both ends by the circumference of a circle is called a chord of the circle.

Def. Every chord through the centre is called a diameter.

Def. Any part of the circumference of a circle is called an arc of the circle.

Def. A proposition which is proved as directly subsidiary to another proposition is called a lemma.

Proposition 1.

PROBLEM--To find the centre of a given circle.

Lemma-Any pt. P equidistant from two given pts. A and B, must lie in the to their join through its mid pt.

B

M

A

Now take AB, AC two chds. of the given O.

Then the centre of the O, being equidistant from A and B, must (by the Lemma) lie in the

through its mid pt. M.

to AB

Similarly it must lie in the L to AC through its mid pt. N.
But these 18 will intersect in some pt. O,

THEOREM-If two points are taken on the circumference of a circle, the chord which joins them must lie within the circle.

i. e. the dist. of P from the centre < the radius.

Similarly every other pt. in AB lies within the O.