A History of Mathematics From Mesopotamia to Modernity

74 A History ofMathematics

K E

A

Z

D

H

B C

L

F

Fig. 9Figure for Heron’s theorem, Appendix A.

[This is the point at which we stop being able to think geometrically; the formula is that

CF.EH=

√

CF^2 ·EH^2

But while the left hand side is a rectangle, the right has arisen by taking the product of two squares
and extracting the square root. From now on, numbers whose ‘dimension’ is 4 seem to come into
the calculation.]
So the area of the triangleABCby itself is the area of the [square] onFCby the [square] onEH.
LetHLbe drawn perpendicular toCHandBLperpendicular toCB, and letCLbe joined. Then
since each of the anglesCHL,CBLis right, a circle can be described about the quadrilateralCHBL
[Euclid III.31]; therefore the anglesCHB,CLBare together equal to two right angles [Eucl. III.22].
But the anglesCHB,AHDare together equal to two right angles because the angles atHare bisected
by the linesAH,BH,CHand the anglesCHB,AHDtogether withAHC,DHBare equal to four right
angles; therefore the angleAHDis equal to the angleCLB. But the right angleADHis equal to the
right angleCBL; therefore the triangleAHDis similar to the triangleCBL.
So asBCis toBL,ADis toDH,thatisasBFis toEH, and interchanging asCBis toBFso isBL
toEH,thatisasBKtoKEbecauseBLis parallel toEH. And putting together, asCFtoBF,soisBE
toEK.
[Note.This is the rule which the translator, following the medieval Latin use, callscomponendo:
ifa/b=c/d, then(a+b)/b=(c+d)/d(Do you see that this works, and applies to the situation?)]
And so the [square on]CFto the [rectangle]CFbyFBis as theBECto theCEK, that is to the
[square on]EH; for in the right angled triangleEHhas been drawn perpendicular to the base.
Therefore, the [square] onCFtimes the [square] onEH, whose side [square root] is the area of the
triangleABC, is equal to theCFBtimes theCEB. And each ofCF,FB,BE,CEis given; forCFis half
the perimeter of the triangleABC, andBFis the excess, by which half the perimeter exceedsCB,
andBEis the excess, by which half the perimeter exceedsAC, andECis the excess, by which half
the perimeter exceedsAB, sinceECis equal toCZ, andBFtoAZ, since it is equal toAD. So the area
of the triangle is given.

Exercise 9.Check through the calculation of the square root of 720. What method is Heron using for

finding it?