1. Let AB be the given straight line on which the isosceles triangle is to be constructed; let DE be the straight line to which each side is to be

With centre A, and radius equal to DE, describe a circle; with Band radius equal to DE, describe another circle; let these circles at C. Join AC and BC; then ABC will be the triangle required. The given point and the vertex of the constructed triangle both fall on the circumference of the small circle.

3. Let AB and CD be two straight lines which bisect each other at right angles at the point 0; so that 40 is equal to OB, CO is equal to OD, and the angles at O are right angles. In CD take any point E, and join EA anl EB: then EA shall be equal to EB.

For AO is equal to BO by hypothesis; EO is common to the two triangles 40E and BOE; and the angle AOE is equal to the angle BOE by Axiom 11. Therefore EA is equal to EB, by I. 4.

Similarly it may be shewn that any point in AB is equally distant from C and D.

4. The angles ABC and ACB are equal by I. 5. Hence the angles DBC and DCB are equal by Axiom 7. Therefore the sides DB and DC are equal by I. 6.

5. The angle DBA is half the angle ABC, by construction. The angle BAD is equal to half the angle ABC, by hypothesis. Therefore the angle DBA is equal to the angle BAD. Therefore BD is equal to AD by I. 6.

6. It is shewn in the demonstration of I. 5, that the angle BCF is equal to the angle CBG; therefore BH is equal to CH, by I. 6. Also it is shewn that FC is equal to GB. Therefore FH is equal to GH by Axiom 3.

7. AF is equal to AG, by construction; AH is common to the two triangles FAH and GAH; and FH is equal to GH, by Exercise 6: therefore the angle FAH is equal to the angle GAH by I. 8.

8. AB is equal to AD, by hypothesis; AC is common to the two triangles BAC, DAC; and the angle BAC is equal to the angle DAC, by hypothesis: therefore the base BC is equal to the base DC, and the angle ACB is equal to the angle ACD by I. 4.

T. EX.

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9. The angle ACB is equal to the angle BDA by I. 8; and then the two triangles ACB and BDA are equal in all respects by I. 4; so that the angle ABC is equal to the angle BAD. Therefore AO is equal to BO by I. 6.

10. Let ABCD be the rhombus, so that AB, BC, CD, DA are all equal. Join BD. Then in the two triangles BAD, BCD the base BD is common; and the two sides BA, AD are equal to the two sides BC, CD, each to each; therefore the angle BAD is equal to the angle BCD, by I. 8. Similarly it may be shewn that the angle ABC is equal to the angle ADC.

11. Let ABCD be the rhombus, so that AB, BC, CD, DA, are all equal. Join BD. Then in the two triangles ABD, CBD, the side AB is equal to the side CB, the side BD is common, and the base AD is equal to the base CD; therefore the angle ABD is equal to the angle CBD by I. 8. Thus the angle ABC is bisected by BD. Similarly it may be shewn that the angle ADC is bisected by BD; and also that the angles BAD and BCD are bisected by AC.

12. Let there be two isosceles triangles ACB, ADB on the same base AB, and on opposite sides of it. Join CD; then CD shall bisect AB at right angles.

In the two triangles ACD, BCD the side CD is common; AC is equal to BC, by hypothesis; and the base AD is equal to the base BD, by hypothesis; therefore the angle ACD is equal to the angle BCD by I. 8.

Let AB and CD intersect at E. Then in the triangles ACE, BCE the side CE is common; the side AC is equal to the side BC; and the angle ACE has been shewn equal to the angle BCE: therefore the triangles are equal in all respects by I. 4. Thus AE is equal to BE; and the angle AEC is equal to the angle BEC, so that each of them is a right angle.

Next let the two isosceles triangles ACB, ADB be on the same base AB, and on the same side of it. Join CD and produce it to meet AB at E. It may be shewn as before that AE is equal to BE, and that the angles at E are right angles.

13. Let AB be the given straight line, C and D the two given points, Join CD and bisect it at E. From E draw a straight line at right angles to CD, meeting AB at F. Join CF, DF. Then CF shall be equal to DF. For in the two triangles CEF and DEF, the side EF is common, CF is equal to DE, and the right angle CEF is equal to the right angle DEF: therefore CF is equal to DF, by I. 4.

The problem is impossible when the two points C and D are situated on the same perpendicular to the given straight line AB, and at unequal distances from that straight line.

14. Let AB be the given straight line, C and D the two given points. From D draw DE perpendicular to AB, and produce DE through E to a point F such that EF=ED. Join CF, and produce it to meet AB. at G; join DG: then CG and DG shall be the required straight lines.

For ED is equal to EF, and EG is common to the two triangles EDG and EFG; the right angles GEF and GED are equal: therefore by I. 4 the triangles FEG and DEG are equal in all respects, so that the angle FGE is equal to the angle DGE.

The problem is impossible when the two points C and D are equally distant from the straight line AB, and not on the same perpendicular to AB.