# Notes of Lessons: Geometry, Class IV

*tour de force*, but is always an illustration or an expansion of some part of the children’s regular studies (in the

*Parents’ Review*School), some passage in one or other of their school books.—Ed.]

Group: Mathematics • Class IV • Age: 16 1/2 • Time: 30 minutes

**By Ida E. Fischer**

*The Parents’ Review, *1903, pp. 144-145

*Introduction to the Second Book of Euclid.*

#### Objects

I. To cultivate the power of inductive reasoning.

II. To give a first introduction to the second book of Euclid, showing its close connection with part of the first book.

III. To teach the first proposition of Book II.

#### Lesson

*Step I*.—Begin by noticing that there is more than one way of proving that figures are equal in area. When they are similar as well as equal the equality may be proved by superposition as in I. 4. This the pupils know and tell themselves. But this cannot be done when we have a parallelogram and a triangle to compare, or two dissimilar triangles or quadrilaterals, for example in Propositions 41 and 43 of Book I. We notice that Book II. consists almost entirely of comparing quadrilaterals and might be taken in connection with parts of Book I.

*Step II*.—Before beginning the proposition show the pupils why rectangles are said to be contained by any two of their conterminous sides. Thus ABCD is said to be contained by AB and BC and it is written AB ∙ BC. The point really being = x. Thus, if AB = 3 and BC = 2, AB ∙ BC = 6.

The area of ABCD is really its length multiplied by its breadth, that is AB multiplied by BC. All this the pupils can give themselves as well as the meaning and derivation of conterminous.

*Step III*.—Begin with the particular enunciation of Proposition I, Book II.

*Particular Enunciation*. Let AB and CD be two given straight lines, let CD be divided into any number of segments, CE, EF, FD.

It is required to prove AB ∙ CD = AB ∙ CE + AB ∙ EF + AB ∙ FD.

*Construction*. From C draw CG ⊥ to CD and = AB. (I. ii. 3.)

Through G draw GH ∥ to CD,

and through EF and D draw EK, FL, DH ∥ to CG. (I. 31.)

*Proof*. Then CH = CK + EL + FH. (I. ax. 8.)

That is GC ∙ CD = GC ∙ CE + KE ∙ EF + LF ∙ FD.

But GC, KE, LF are each = AB. (Constr. I. 34.)

**∴** AB ∙ CD = AB ∙ CE + AB ∙ EF + AB.

Let the girls do as much of the work as possible without help, such as constructing the figure and giving the proof.

*Step IV*.—Let the girls write out the proof, having the figure on the board, and from the proof let them give the general enunciation: —“If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangle contained by the undivided line and the several parts of the divided line.” Do not have the long enunciation learnt by heart.

*Step V*.—Recapitulate.