P l a n e   G e o m e t r y

An Adventure in Language and Logic

based on

Book I.  Propositions 33 and 34

Proposition 33

WE ARE MAKING OUR final approach to the theorem of Pythagoras. But we first have to establish when figures that are not congruent will be equal. To do this, we will look at quadrilaterals whose opposite sides are parallel. Such a figure is called a parallelogram. (Definition 15.) The following proposition effectively shows that such a figure exists.

PROPOSITION33.  THEOREM

The proposition proves that if two sides of a quadrilateral are equal and parallel, then the figure is a parallelogram. (Definition 14.) Hence we may construct a parallelogram; for, Proposition 31 shows how to construct a straight line parallel to a given straight line.

The next theorem has for its hypothesis that a figure is a parallelogram, that is, the opposite sides are parallel. And it proves what is obvious to the eye:  the opposite sides are equal.

This theorem will be fundamental to the theory of equal areas. In fact, as we have pointed out several times, when we say that two figures are "equal," we mean that they are equal areas. See Problems 1 and 2 following Proposition 4.

The student should look at the figure below and, given that the figure is a parallelogram, it should be clear why those two triangles are equal.

Why?  S.A.S.

PROPOSITION34.  THEOREM

 In a parallelogram the opposite sides and angles are equal, and the diagonal bisects the area. Let ACDB be a parallelogram, and BC its diagonal; then the opposite sides and angles are equal, and the diagonal BC divides the parallelogram into two equal areas. Since AB is parallel to CD, and CB meets them, the alternate angles ABC, BCD are equal; (I. 29) and since AC is parallel to BD, and CB meeets them, the alternate angles ACB, CBD are equal. In triangles ABC, DCB, then, the two angles ABC, BCA are equal to the two angles DCB, CBD respectively,and the side BC is common; therefore the remaining sides will equal the remaining sides: AB will equal CD, and AC will equal BD; and the remaining angle will equal the remaining angle: angle CAB will equal angle BDC. (A.S.A.) Also, because angle ABC is equal to angle BCD, and angle ACB is equal to angle CBD, the whole angle ABD is equal to the whole angle ACD. (Axiom 2) Therefore the opposite sides and angles of a parallelogram are equal. Next, the diagonal bisects the parallelogram. For, since AB is equal to CD, and BC is common, the two sides AB, BC are equal to the two sides DC, CB respectively; and we have shown that angle ABC is equal to angle BCD; therefore triangles ABC, DCB are equal. (S.A.S.) Therefore the diagonal BC divides the parallelogram into two equal areas. Therefore, in a parallelogram etc.  Q.E.D.

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