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.. rect. AD, DB, together with sq. on CD = sq. on CB.

Therefore, if a straight line, &c. Q. E. D. Cor. From this proposition it follows, that the difference

of the squares on two unequal lines is equal to the rectangle contained by their sum and their difference.

[The corollary follows thus :— Let AC, CD be two lines of which AC is the greater ; produce AC to B, making CB = AC. Then, ... AB is divided into two equal parts in C and into two unequal parts in D, rect. AD, DB together with sq. on CD sq. on AC; take away from each of these =s sq. on CD, and rect. AD, DB = difference of sqs. on AC and CD.

Algebraically. Let AB contain 2a linear units, and CD b units. Then AD contains a +b, and DB a-6 units; now (a + b)(a-6)= a -6: add bato each, and (a + b)(a-6) + b2 = a”; that is, if a number be divided into two equal parts, &c.]

AD is drawn from the vertex of an isosceles triangle to any point D in the base BC. Show that the difference of the squares on AB, AD is equal to the rectangle contained by BD, CD.

PROPOSITION VI. THEOREM.

If a straight line be bisected and produced to any point, the rectangle contained by the whole line thus produced and the part of it produced, together with the square on half the line bisected, is equal to the square on the line which is made up of the half and the part produced.

Let AB be bisected in C and produced to D.
Then rect. AD, DB, together with sq. on BC, = sq.

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Constr. On CD describe sq. CF, (i. 46.) and join DE,

through B draw BHG || CE or DF, through H draw KLM || AD or EF, and through A draw AK || CL or DM.

(i. 31) Dem. Then, :: AC= CB, .-. AL = CH; (i. 36) but CH=HF; (i. 43).. AL=HF; to each add CM;

.. AM = gnomon CMG : but AM is rect. AD, DB, for DM = DB; .. gnomon CMG = rect. AD, DB : to each add LG, which = sq. on CB;

(ii. 4, cor.) .. rect. AD, DB with sq. on CB = gnomon CMG and

but gnomon CMG and LG make up the whole figure CF, which is sq. on CD;

LG;

.. rect. AD, DB, together with sq. on CB = sq. on CD.

Therefore, if a straight line, &c. Q. E. D.

[The corresponding result is thus shown algebraically :Let AB contain 2a linear units, then BC .contains a units; let BD contain x units, then AD contains 2a + x, and since x(2a + x) = 2ax + x*; .. (adding aż to each), a2 + x(2a + x) =a? + 2ax + x, = (a + x)"; that is, if a number be divided, &c.]

PROPOSITION VII. THEOREM.

If a straight line be divided into any two parts, the squares on the whole line, and on one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square on the other part.

Let AB be divided into any two parts at C.

Then the sqs. on AB, BC =twice rect. AB, BC together with sq. on AC.

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Constr. On AB describe sq. AE, (i. 46) join BD, through

C draw CF || AD or BE, cutting BD at G, and through G draw HGK || AB or DE.

(i. 31) Dem. Then, :: AG=GE, (i. 43) add CK to each ;

.. AK = CE, and .. AK, CE are double of AK; but AK, CE are the gnomon AKF and sq. CK;

.. gnomon AKF and sq. CK are double of AK ; but twice rect. AB, BC is double of AK, for BK = = BC; i. gnomon AKF and sq. CK = twice rect. AB, BC; add to each HF, which = sq. on AC;

(ii. 4, cor.) .. gnomon AKF, and sqs. CK, HF = twice rect. AB, BC

and sq. on AC; but gnomon AKF and sqs. CK, HF make up the whole

figure AE and CK, which are the sqs. on AB, BC;

.. sqs. on AB and BC = twice rect. AB, BC together with sq. on AC. Therefore, if a straight line, &c.

Q. E. D.

[Algebraical form :-Let AB contain a units, and let it be divided into any two parts at C; let AC contain x units ; then BC contains a-x units. Now (a - x)2 = a? — 2ax + x*, .. a? + (a — x)2 = 2a? – 2ax + x2 = 2a(a – x) +22; that is, if a number be divided into any two parts, &c.]

1. From the right angle of a right-angled triangle a perpendicular is drawn on the hypotenuse: show that the square on each of the sides is equal to the rectangle contained by the hypotenuse and the segment of it adjacent to that side.

2. If a perpendicular be drawn from the vertical angle of a triangle on the base, and the square on each side be equal to the rectangle contained by the base and the segment of it adjacent to that side, then the vertical angle is a right angle.

3. Show that the sum of the squares on two straight lines is never less than twice their rectangle.

4. Produce a given line so that the rectangle of the whole line produced and the given line shall be equal to a given square.

Is this always possible?

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