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A history of Greek mathematics - Wilbourhall.org

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one<br />

:<br />

THE COLLECTION. BOOK VII 421<br />

in K, L; and draw LAI parallel to AD meeting GE produced<br />

in M.<br />

HE.FG HE FG LH A£ LH<br />

n<br />

'<br />

EG . EF~ EF EG ~ '<br />

AF HK == EK<br />

'<br />

In exactly the same way, if BE produced meets LM in M'<br />

we prove that<br />

Therefore<br />

(The proposition is<br />

HB . CD LE<br />

ED.BC EK<br />

EE.FG EB.CD<br />

EG.EF" ED.BC<br />

proved for IfBCD and any other transversal<br />

not passing through E by applying our proposition<br />

twice, as usual.)'<br />

Props. 136, 142 are the reciprocal; Prop. 137 is a particular<br />

case in which one <strong>of</strong> the transversals is parallel to one <strong>of</strong> the<br />

straight lines, Prop. 140 a reciprocal <strong>of</strong> Prop. 137, Prop. 145<br />

another case <strong>of</strong> Prop. 129.<br />

The Lemmas 12, 13, 15, 17 (Props. 138, 139, 141, 143) are<br />

equivalent to the property <strong>of</strong> the hexagon inscribed in two<br />

straight lines, viz. that, if the vertices <strong>of</strong> a hexagon are<br />

situate, three and three, on two straight lines, the points <strong>of</strong><br />

concourse <strong>of</strong> opposite sides are in a straight line ; in Props.<br />

138, 141 the straight lines are parallel, in Props. 139, 143 not<br />

parallel.<br />

Lemmas 20,<br />

21 (Props. 146, 147) prove that, when one angle<br />

<strong>of</strong> one triangle is equal or supplementary to<br />

•<br />

angle <strong>of</strong><br />

another triangle, the areas <strong>of</strong> the triangles are in the ratios<br />

<strong>of</strong> the rectangles contained by the sides containing the equal<br />

or supplementary angles.<br />

The seven Lemmas 22, 23, 24, 25, 26, 27, 34 (Props. 148-53<br />

and 160) are propositions relating to the segments <strong>of</strong> a straight<br />

line on which two intermediate points are marked. Thus<br />

Props. 148, 150.<br />

If C<br />

f<br />

D be two points on AB, then<br />

(a) if 2AB.CD = CB 2 , AD<br />

Z ^AC 2 + DB 2 \<br />

A<br />

i—<br />

C<br />

1<br />

D<br />

1<br />

B*<br />

(b) if 2AC.BD = CD*, AB 2 = AD 2 + CB\

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