arc BH is added, and the angle DBC is equal to the angle DAG, each being measured by half the arc DC. Hence the two remaining angles BCD and AGD are equal. Therefore

and hence

AD: DB:: AG: BC;

ADX BC=DBX AG.

By adding the two equations,

ABXDC+AD × BC=GC × DB+AG×DB; AB×DC+AD ×BC=(GC+AG) × DB, or=AC × DB.

The following are Test Examples involving Book Fifth:

1. To bisect a quadrilateral by a line drawn from one of its angles.

Α

D

H

B

Let DCB be the angle from which the line shall be drawn bisecting the given quadrilateral ABCD.

Draw the diagonals DB and AC; bisect DB in the point E, and through E draw FEG

parallel to AC. Join AE and EC, and from C draw CF. CF bisects the quadrilateral.

The triangles DEC and BEC are equal (Prop. VIII., Bk. II.), and also AED and AEB, for the same reason. By adding AED and EDC, and AEB and ECB, two by two, the quadrilateral AECD will be equal to the quadrilateral AECB. The triangles AFE and FEC are equal (Prop. VIII., Bk. II.). From each take away the common triangle FHE, and there will remain AFH equal to HEC. Now if from the quadrilateral AECD the triangle HEC be taken away, and its equal AFH added, we shall have the quadrilateral AFCD; and if from AECB AFH be taken away, and its equal HEC added, we shall have the

triangle FCB. Hence FCB is equal to AFCD, and FC bisects the quadrilateral.

2. To determine the figure formed by joining the points of bisection of the sides of a trapezium, and its ratio to the trapezium.

For, since AD and DC are bisected in G and F, AD: GD:: DC: DF, and hence GF is parallel to AC; and in the same manner EH is parallel to AC. Therefore GF and EH are parallel to each other. In like manner, it can be proved that FH and GE are parallel to each other. Hence GEHF is a parallelogram.

Again, the triangle ADC is composed of the triangles ALG, GND, DNF, FKC (without the parallelogram), and of the triangles GLN, LNO, ONK and NFK (within the parallelogram). Now it is evident that ALG and GND are equal to LNG and LNO, and also that DNF and FKC are equal to NFK and ONK. By adding these, it will be found that the parallelogram GLKF is equal to the sum of the triangles ALG, GND, NDF, and FKC. In the same manner, it can be proved that the parallelogram LEHK is equal to the sum of the triangles ALE, EMB, MBH, and HKC. Hence the whole parallelogram is one half the trapezium.

3. If from any point in the base of an isosceles triangle perpendiculars be drawn to the sides, these together shall be equal to a perpendicular drawn from either extremity of the base to the opposite side.

Let ABC be an isosceles triangle; and from any point,

B

D

G

C

D, draw the perpendiculars DE and DF; also the perpendicular BG. It is required to prove that BG is equal to the sum of DE and DF.

In the two triangles BED and DCF, the angles EBD and C are Fequal (hyp.), and BED and DFC are also equal, being right angles. Hence the triangles are similar, and BD: DC:: DE: DF; and, by composition, BD+DC, or BC: DE +DF:: DC: DF. But BG being parallel to DF, DC: DF:: BC: BG, whence BC: BG:: BC: DE+DF. Therefore BG is equal to DE+DF.

TEST EXAMPLES IN BOOK V.

1. Through a given point situated between the sides of an angle, to draw a line terminating at the sides of the angle, and in such a manner as to be equally divided at the point.

2. Construct a quadrilateral similar to a given quadrilateral, the sides of the latter having to the sides of the former the ratio of 2 to 3.

3. To draw a line parallel to the base of a triangle in such a manner as to divide the triangle into two equal parts. 4. To construct a square when the difference between the diagonal and a side is given.

5. Prove that, if a line touching two circles cut another line joining their centres, the segments of the latter will be to each other as the diameters of the circles.

6. Prove that, if from the extremities of any chord in a circle, perpendiculars be drawn meeting a diameter, the points of intersection are equally distant from the centre.

7. Prove that, if from the extremities of the diameter of a semicircle, perpendiculars be let fall on any line cutting the semicircle, the parts intercepted between those perpendiculars and the circumference are equal.

8. Prove that if two circles touch each other externally or internally, any straight line drawn through the point of contact will cut off similar segments.

9. To determine the point in the circumference of a circle from which lines drawn to two other given points also in the circumference shall have a given ratio.

10. Prove that in any right-angled triangle the straight line joining the right angle and the point of bisection of the hypothenuse is equal to half the hypothenuse.

11. To construct a polygon similar to a given polygon, and bearing to it a given ratio.

12. Prove that, if from any point within an equilateral triangle, perpendiculars be drawn to the sides, they are together equal to a perpendicular drawn from any of the angles to the opposite side.

13. Prove that if the points of bisection of the sides of a given triangle be joined, the triangle so formed will be one-fourth the given triangle.

14. Prove that of all triangles having the same vertical angle and whose bases pass through a given point, the least is that whose base is bisected in the given point.

15. To bisect a given triangle by a line drawn from one of its angles.

16. To bisect a given triangle by a line drawn from a given point in one of its sides.

17. To determine a point within a given triangle from which lines drawn to the several angles will divide the triangle into three equal parts.

18. To trisect a given triangle from a given point within it.

19. Prove that if the three sides of a triangle be bisected, the perpendiculars drawn to the sides at the three points of bisection will meet in the same point.

20. Prove that if, from the three angles of a triangle, lines be drawn to the points of bisection of the opposite sides, these lines intersect each other in the same point.

21. Prove that every trapezoid is divided by its diagonals into four triangles proportional to each other.

22. To describe a triangle which shall be equal to a given equilateral and equiangular pentagon, and of the same altitude.

23. Prove that if an equilateral triangle be inscribed in a circle, and through the angular points another be circumscribed, the inscribed will be one-fourth the circumscribed.

24. Prove that the ratio of the side of a square to the diagonal is that of 1 to the √2.

C D

A KEY TO THE TEST EXAMPLES IN BOOK V.

A

1. A is the given angle, and O the given point. Join AO, and produce it until OD is equal to AO; then through D draw DF parallel to AB. The line FE is the line required. The student will have no difficulty in proving FO=OE.

2. Trisect the sides of the former: two of these parts respectively will be the sides of the latter: then see Prop. XIV., Bk. I. 3. See Prop. XVII. The triangle is given; also its sides. The student will readily discover the proportion whose fourth term will be the portion of the side through the extremity of which the line must be drawn parallel to the base.

14

E

B

D

4. Let AB be the given difference: at the point A make DAB right angle; at B make ABD 14 right angles: produce these lines until they meet: at D make BDC=right angle: produce DC until it meets AB prolonged. Through C draw EC parallel to AD, and AE, through A, parallel to DC.

The student can now readily prove DE a square, and

BC to one side.

angles ACE and EBD.

5. AB is tangent to both circles: CD joins their centres. The student will have little difficulty in proving the similarity of the tri