PROPOSITION 5. THEOREM.

If a straight line is divided equally and also unequally, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line.

Let the straight line AB be divided equally at P, and unequally at Q.

Then the rect. AQ, QB together with the sq. on PQ shall be equal to the sq. on PB.

Construction.

On PB describe the square PCDB. I. 46.

Join BC.

Through Q draw QE par' to BD, cutting BC in F. I. 31.
Through F draw LFHG par' to AB.

Proof.

Through A draw AG par1 to BD.

=

Now the complement PF the complement FD:

to each add the fig. QL;

then the fig. PL = the fig. QD.

I. 43.

But the fig. PL = the fig. AH, for they are parts on equal bases and between the same parls;

.. the fig. AH = the fig. QD.

To each add the fig. PF;

then the fig. AF = the gnomon PLE.

Now the fig. AF is the rect. AQ, QB; for QF = QB ;
the rect. AQ, QB = the gnomon PLE.

I. 36.

To each add the sq. on PQ, that is, the fig. HE; II. 4. then the rect. AQ, QB with the sq. on PQ

gnomon PLE with the fig. HE the whole fig. PD,

which is the sq. on PB.

That is, the rect. AQ, QB together with the square on PQ is equal to the sq. on PB.

Q.E.D.

COROLLARY. From this Proposition it follows that the difference of the squares on two straight lines is equal to the rectangle contained by their sum and difference.

For let X and Y be the given A

st. lines, of which X is the greater. Draw AP equal to X, and produce it to B, making PB equal to AP, that is to X.

X

X-Y

P

B

From PB cut off PQ equal to Y.

Then AQ is equal to the sum of X and Y,

and QB is equal to the difference of X and Y.

Now because AB is divided equally at P and unequally at Q, the rect. AQ, QB with sq. on PQ=the sq. on PB; II. 5. that is, the difference of the sqq. on PB, PQ= the rect. AQ, QB. or, the difference of the sqq. on X and Y = the rectangle contained by the sum and the difference of X and Y.

CORRESPONDING ALGEBRAICAL FORMULA.

This result may be written

AQ. QB+PQ2= PB2.

Let AB=2a; and let PQ=b;

then AP and PB each=a.

Also AQ=a+b; and QB-a - b.

Hence the statement AQ. QB + PQ2= PB2

In the above figure shew that AP is half the sum of AQ and QB; and that PQ is half their difference.

If a straight line is 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 straight line made up of the half and the part produced.

Let the straight line AB be bisected at P, and produced to Q.

Then the rect. AQ, QB together with the sq. on PB shall be equal to the sq. on PQ.

Construction. On PQ describe the square PCDQ. 1. 46.

Join QC.

Through B draw BE par1 to QD, meeting QC in F.
Through F draw LFHG par1 to AQ.
Through A draw AG par1 to QD.

I. 31.

Proof. Now the complement PF = the complement FD. I. 43. But the fig. PF = the fig. AH; for they are parts on equal bases and between the same parls.

.. the fig. AH = the fig. FD.

To each add the fig. PL;

then the fig. AL = the gnomon PLE.

Now the fig. AL is the rect. AQ, QB; for QL=QB ; .. the rect. AQ, QB = the gnomon PLE. To each add the sq. on PB, that is, the fig. HE; then the rect. AQ, QB with the sq. on PB

I. 36.

= the gnomon PLE with the fig. HE the whole fig. PD,

=

which is the square on PQ.

That is, the rect. AQ, QB together with the sq. on PB is

equal to the sq. on PQ.

Q.E.D.

Also AQ=a+b; and QB=b-a.

Hence the statement AQ. QB + PB2=PQ2

DEFINITION. If a point X is taken in a straight line AB, or in

AB produced, the distances of

the point of section from the A extremities of AB are said to be

divided at X.

A

the segments into which AB is

In the former case AB is

divided internally, in the latter case externally.

Thus in each of the annexed figures, the segments into which AB is divided at X are the lines AX and XB.

This definition enables us to include Props. 5 and 6 in a single Enunciation.

If a straight line is bisected, and also divided (internally or externally) into two unequal segments, the rectangle contained by the unequal segments is equal to the difference of the squares on half the line, and on the line between the points of section.

EXERCISE.

Shew that the Enunciations of Props. 5 and 6 may take the following form:

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

[See Ex., p. 137.]

PROPOSITION 7. THEOREM.

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

Let the straight line AB be divided at C into the two parts AC, CB.

Then shall the sum of the sqq. on AB, BC be equal to twice the rect. AB, BC together with the sq. on AC.

Construction.

On AB describe the square ADEB.

Join BD.

I. 46.

Through C draw CF par1 to BE, meeting BD in G. I. 31.
Through G draw HGK par' to AB.

Proof. Now the complement AG = the complement GE; I. 43. to each add the fig. CK:

then the fig. AK = the fig. CE.

But the fig. AK is the rect. AB, BC; for BK BC;

=

the two figs AK, CE = twice the rect. AB, BC. But the two figs. AK, CE make up the gnomon AKF and the fig. CK:

... the gnomon AKF with the fig. CK = twice the rect. AB, BC. To each add the fig. HF, which is the sq. on AC:

then the gnomon AKF with the figs. CK, HF

=twice the rect. AB, BC with the sq. on AC. But the gnomon AKF with the figs. CK, HF make up the figs. AE, CK, that is to say, the sqq. on AB, BC;

.. the sqq. on AB, BC= twice the rect. AB, BC with the sq. on AC.

Q.E.D.