Inhaltsangabe
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 equal. With centre A, and radius equal to DE, describe a circle; with centre Band radius equal to DE, describe another circle; let these circles intersect at C. Join AC and BC; then ABC will be the triangle required. 2. 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 AO is equal to OB, CO is equal to OD, and the angles at Oare right angles. In CD take any point E, and join EA and EB :then EA shall be equal to EB. For AO is equal to BO by hypothesis; EO is common to the two triangles AOE 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 Cand 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 Cll, by I. 6. Also it is shuwn that EC is equal to GB. Therefore EH 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, byE xercise 6: therefore the angle FAH is equal to the angle GAH by I. 8. 8. AB is equal to AD, by hypothesis; AG is common to the two triangles BAG, DAC; and the angle BACis equal to the angle DAC, by hypothesis :therefore the base B
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Excerpt from Key to Exercises in Euclid
Next let the two isosceles triangles agb, 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 AB is equal to be, and that the angles at E are right angles.
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