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3. That if two As have two sides and the between them, in the one, equal respectively to two sides and the between them in the other, those As are also equal in every other respect. (Prop. IV.)

4. That if a parallelogram and a ▲ be upon the same base and between the same parallels, the parallelogram is double the A. (Prop. XLI.)

Let A C B be a ▲ in which the

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ACB is a

.

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On the three sides

describe the

squares A DE B,

I ACGF and BIH C. We have to prove that the square ADEB is equal to the sum of the other two squares A C G F and BIH C.

From the point C draw the line C M parallel to AD (or

B E), and meeting the line in D E in the point M, and cutting A B in the point L.

Join the points F and B by the straight line F B; C and D by the straight line CD; C and E by the straight line CE; and A and I by the straight line A. I.

The mode in which the proposition is proved is, to show that the square ACGF is equal to the parallelogram ALMD, and the square CBIH to the parallelogram LBEM; so that the sum of the two squares will be equal to the sum of the two rectangles, i.e., to the square ABED.

To prove that the square ACGF is equal to the parallelogram ALMD, it is shown-1, that the square

ACGM is double of the ▲ FAB; 2, that the parallelogram A LMD is double of the DAC; 3, that theses are equal to one another; 4, that their doubles are also equal. In a similar manner it is proved, 5, that the square B CHI is double of the ▲ ABI; 6, that the parallelogram LBEM is double of the ▲ ECB; 7, that the As A BI and ECB are equal to one another; and 8, that their doubles are also equal to one another.

1. The square A C G F is double of the ▲ FAB. Since G C A is an of a square, it is a right : the ACB is also a right . Consequently, we have two straight lines, G C and B C, meeting a third, A C, at the same point C, and on opposite sides, and making the two adjacent ≤s GCA and BCA together equal to two right ▲s.

It follows (according to the 14th proposition), that the two lines G C and CB form one continued straight line. And since the opposite sides of a square are parallel, the line G B is parallel to the line F A.*

Hence it appears that the square (or parallelogram) ACGF and the AFAB are upon the same base, FA; and between the same parallels, G B and F A. Therefore the square AC G F is double the A FAB.

2. The parallelogram D M LA is double the A DAC; for CM was drawn parallel to A D, and consequently the parallelogram ADML and the A CAD are upon the same base, AD; and between the same parallels, A D and CM.

* Beginners are very apt to omit or misunderstand this part of the proof. They are prone to look upon it as a part of the construction that GCB is made one continued straight line, and to forget that it is made up of two separate lines, which must be proved to be one continued right line, before we are entitled to affirm that the whole of GCB is parallel to F A.

K

Therefore the parallelogram ADML is double the A DAC.

3. The ▲ FAB is equal to the ▲ DA C.

For the lines FA and A C are equal, being sides of the same square. The lines A B and A D are also equal, being sides of the same square. Moreover, since each of the s FAC and D A B is an of a square, they are rights, and therefore equal to one another. Add to each of them the CAB; and the sum of the s FAC and CAB (i.e., the FA B) will be equal to the sum of the s DAB and C A B (i.e.,. the DAC).

Hence the two sides F A and AB and the FAB in the ▲ FA B, are equal respectively to the two sides CA and AD and the CAD in the ▲ CAD. Consequently (according to the 4th proposition), these two As are equal in every respect and among others, in area.

4. Since the doubles of equals are equal (Ax. VII.), the square F A C G (which is double the ▲ FA B) is equal to the parallelogram A LMD (which is double. the ▲ CAD).

We have next to show that the square C BIH is equal to the parallelogram LBE M.

The proof of this part of the demonstration is of the same kind as that of the other.

The HCB is a right, being an ▲ of a square. The ACB is also a right

.

Therefore, since the two lines AC and HC meet the line BC at the same point and on opposite sides, and make the adjacents together equal to two right <s, those two lines lie in one and the same straight line (Prop. XIV.); and HA is parallel to BI.

Hence it appears that the square BIHC and the A BIA are upon the same base and between the same parallels.

Therefore the square BCHI is double the A BIA.

Again, the parallelogram L BEM and the ▲ EBC are upon the same base, BE; and between the same parallels, BE and CM. Consequently, the parallelogram L BE M is double of the ▲ EBC.

But the As ABI and EBC are equal to one another.

For the lines BI and BC are equal, being sides of the same square.

Similarly, the lines B E and B A are equal, being sides of the same square. And since the Zs CBI and ABE are both rights, and therefore equal to one another, if the CBA be added to both, the sums are equal. That is to say, the ABI is equal to the EB C.

Hence it appears, that the sides A B and B I, and the included ABI in the one ▲, are equal respectively to the sides E B and B C, and the included EBC in the other ▲.

Therefore, the A ABI is equal to the ▲ EBC. (Prop. IV.)

And since the doubles of equals are equal, it follows, that the square CBIH is equal to the parallelogram LBEM.

But if equals be added to equals, the sums are equal (Ax. II.)

Therefore the sum of the squares AC G F and BCHI is equal to the sum of the parallelograms ALMD and LBEM; that is, to the whole square A BED.

PROPOSITION XLVIII.

If the square described upon one of the sides of a triangle be equal to the sum of the squares described upon the other two sides, the angle contained by these two sides is a right angle.

For the construction employed in this proposition we must be able,—

1. To join two given points by a straight line. (Post. I.)

2. From a given point in a given right line to draw a right line perpendicular to that line. (Prop. II.)

To prove the proposition by the aid of the construction we must know,

1. That if the same be added to equals, the sums are equal. (Ax. II.)

2. That things that are equal to the same are equal to each other. (Ax. I.)

3. That if one of a ▲ be a right, the square on the side which subtends the right is equal to the sum of the squares on the sides which contain the right. (Prop. XLVII.)

4. That if two As have the three sides of the one equal respectively to the three sides of the other, those As are also equal in every other respect. (Prop. VIII).

D

B

Let ACB be a ▲, such, that, when
squares are described upon the
three sides of it, the square de-
scribed on the side A B is equal to
the sum of the squares described on
the sides AC and C B.

It has to be shown that the
ACB is a right Z.

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