Notes of lessons on the Herbartian method (based on Herbart's plan) . If one is a right angle, what do weknow about the other two ? They must together equal a rightangle. What do we say two angles are to one another whentogether they make two right angles ? Then from Proposition 32 we can make one it. (Write on blackboard.) Make class write out Proposition I. 32. NOTES OF A LESSON ON A RIDER INEUCLID. Class—Average age, 15. Time—Half an hour. Previous Knowledge—Book I. to end of Proposition 32. Aim—To exercise the reasoningpowers of the pupils and train them to accuracy of judg
Notes of lessons on the Herbartian method (based on Herbart's plan) . If one is a right angle, what do weknow about the other two ? They must together equal a rightangle. What do we say two angles are to one another whentogether they make two right angles ? Then from Proposition 32 we can make one it. (Write on blackboard.) Make class write out Proposition I. 32. NOTES OF A LESSON ON A RIDER INEUCLID. Class—Average age, 15. Time—Half an hour. Previous Knowledge—Book I. to end of Proposition 32. Aim—To exercise the reasoningpowers of the pupils and train them to accuracy of judgment by leadingthem to deduce the solution of the following rider. Preparation. Blackboard Illustrations. Given an isosceles triangle ABC. ^Required to draw a straight line DEEnunciation ana-J parallel to BC, meeting the equal lysed. I sides at D and E, so that the lines V DE, DB and EC are all equal. 204 Notes on Herbartian Method II. Presentation. i. Make DE asrequired andwork back-wards to arriveat steps ofconstruction. 2. 3. Proof:proveDE. (a) ToDB = (b) To prove= DB. EC DB then triangle DBE isisosceles, therefore angleDBE « angle DEB (I. 5).But because it is parallel toBC and BE meets them,therefore angle DEB — angleEBC (I. 29). Thereforeangle DBE = angle EBCC (Axiom 1). Bisect angle DBC by BE meeting ACat E (I. 10). Through E draw EDparallel to BC (I. 31). Then shallDE = DB and EC. Because DE is parallel to BC and BEmeets them, angle DEB = angleEBC (I. 29). But angle DBE =angle EBC (Construction), there-fore angle DEB = angle DBE(Axiom 1); therefore triangle BDEis isosceles and DB = DE.(Because DE is parallel to BC and linesAB and AC meet them, angle ADE= angle ABC, and angle AED -angle ACB (I. 29). But because triangle ABC is isosceles,angle ABC = angle ACB (I. 5);therefore angle ADE = angle AED ;therefore AD = AE (I. 6). III. Association. Propositions I. 10, I. 29,1. 5, I. 6, I. 31Definitions, isosceles and parallels. Axioms 1 a
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