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And AE is less than MF+ HP by MP + HF; that is, the square on AB is equal to the squares of AP, PB diminished by twice the rectangle contained by AP, PB.
The difference of the squares on two lines is equal to the rectangle contained by the sum and difference of the lines *.
Part. En. Let AB and BC be the two straight lines, of which AB is the greater; and let them be placed in one straight line; cut off BD equal to BC; so that AC is their sum, and AD is their difference;
Then will the difference of the squares of AB and BC be equal to the rectangle contained by AC and AD.
Proof. On AB describe a square AGHB. Through D, C draw DL, CF parallel to AG or BH; cut off HO= LH or DB; and through O draw EKOF parallel to AC.
Then KH is the square on DB or BC; and therefore the difference of the squares of AB and BC is the figure made up of EL and A0.
But EL is equal to BF by construction; therefore the figure made of EL and AO is equal to AF; which is the rectangle contained by AE or AD and AC;
* Euclid, 11. 5, Cor.
therefore the difference of the squares of AB and BC is equal to the rectangle contained by AC and AD.
Cor. If a straight line is bisected and divided in any point, the rectangle contained by the segments is equal to the difference of the squares on half the line and the line between the points of section.
A Ç P B · C B P
Proof. For let AB be bisected in C, and divided internally or externally in P.
Then AP is the sum of AC and CP, and PB is their difference, since BC= AC.
Therefore the rectangle contained by AP, PB is the rectangle contained by the sum and difference of AC and CP, and therefore is equal to the difference of the squares of AC and CP.
Remark. The student will begin here to suspect, what he will afterwards find to be true, that there is an intimate relation between geometry and algebra. Algebraical or analytical geometry as it is called, investigates this relation and applies it to the establishment of theorems in geometry, and will occupy him at a later stage of his mathematical studies. We shall at present use the expression AB, which is read 'AB squared,' only as an abbreviation for “the square on AB," and ABX AC or AB. AC, as an abbreviation for “the rectangle contained by AB and AC.”
In any right-angled triangle the square on the hypotenuse is equal to the sum of the squares on the sides which contain the right angle.
! Part. En. Let ABC be a
Proof. On AB, BC, CA describe the squares ADEB, BFGC, CIHA respectively. Join CD, BH; and draw BJ parallel to AH.
Since the angles ABC, ABE, BCF are right angles, it follows that CBE, ABF are straight lines; (1. 3.) therefore the triangle DAC is on the same base DA, and be. tween the same parallels DA, EC with the square DABE; therefore the triangle DAC is half the square DABE;
(11. 3, Cor. 2.) and similarly the triangle BAH is half the rectangle AJ.
But because the angle DAB= the angle HAC, each being a right angle; add to each the angle BAC; therefore the whole angle DAC is equal to the whole angle BAH; and the two sides DA, AC are respectively equal to the two sides BA, AH;
(Constr.) therefore the triangle DAC is equal to BAH; (1. 5.) and therefore the square DABE = the rectangle AJ.
Similarly it may be shewn that the square BCGF=the rectangle CJ, and therefore, since AJ and CJ make up
the whole square AHIC, the square AHIC is equal to the sum of the squares ABDE and BCGF, that is,
ACP= ABR + BC%*
Place two squares EABH, KBCF, as in the figure, with their sides AB, BC continuous and in the same straight line. From CB cut off CD equal to AB. Join DE, DF.
Produce BK to L, making KL= AB. Join LE, LF.
Then it will be easy to prove that the triangles EAD, DCF, EHL, LKF are all equal, being right-angled, and having the sides containing the right angle equal; therefore the figure LEDF is equal to the sum of the two given squares, and all its sides are equal.
And since EDA is complementary to AED or FDC, therefore the angle EDF is a right angle. Therefore LEDF is a square, and is the square on ED.
Therefore the square on the hypotenuse ED is equal to the sum of the squares on the sides EA, AD,
* Euclid, 1. 47.
Cor. I. It follows that in a triangle ABC right-angled at B,
ABR = AC? - BC and BC=AC - AB'.
The next two theorems shew the modifications which the theorem undergoes when the triangle is not right-angled.
In an obtuse-angled triangle the square on the side subtending the obtuse angle is greater than the squares on the sides containing that angle by twice the rectangle contained by either of these sides and the projection on it of the other side*.
* Euclid, 11. 12.