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3. That if two As have two sides and the <
between them, in the one, equal respectively to
the same base and between the same parallels, the parallelogram is double the A. (Prop. XLI.)
Let A C B be a A in which the < AC B is a
a right Z.
On the three sides H
of this A describe the squares
A DE B, I ACGF and BIH C.
We have to prove that the square A DEB is equal to the sum of the other two squares ACGF and BIH O.
From the point o draw the line CM
parallel to A D (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 FB; 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 L BEM; 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 A FAB; 2, that the parallelogram ALMD is double of the ADAC; 3, that these As 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 A ABI; 6, that the parallelogram LBEM is double of the AECB; 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 ACGF is double of the A FA B.
Since G C A is an < of a square, it is a right 2: the < ACB is also a right 2. Consequently, we have two straight lines, G C and B C, meeting a third, AC, at the same point C, and on opposite sides, and making the two adjacent Z8 GCA and BCA together equal to two right Zs.
It follows (according to the 14th proposition), that the two lines G C and C B 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 A FAB are upon the same base, FA; and between the same parallels, G B and F A.
Therefore the square ACGF is double the A FAB.
2. The parallelogram DMLA is double the A DAC; for C M was drawn parallel to AD, and consequently the parallelogram A DML 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 G C B 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 G C B is parallel to F A.
Therefore the parallelogram ADML is double the ADA C.
3. The A FAB is equal to the ADAC.
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 8 FAC and D A B is an < of a square, they are right Z8, and therefore equal to one another. Add to each of them the 2 CAB; and the sum of the Zs FAC and C AB (i.e., the ZFAB) will be equal to the sum of the Zs DAB and C A B (i.e., the <D AC).
Hence the two sides F A and AB and the ZFAB in the A F A B, are equal respectively to the two sides C A and A D and the < CAD in the A 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 FACG (which is double the A FAB) is equal to the parallelogram ALMD (which is double the ACAD).
We have next to show that the square C BIH is equal to the parallelogram LBEM.
The proof of this part of the demonstration is of the same kind as that of the other.
The LHCB is a right Z, being an < of a square. The < ACB is also a right Z.
Therefore, since the two lines AC and HC meet the line BC at the same point and on opposite sides, and make the adjacent Zs together equal to two right <s, those two lines lie in one and the same straight line (Prop. XIV.); and H A is parallel to BI.
Hence it appears that the square BIHC and the A BI A are upon the same base and between the same parallels.
Therefore the square BCHI is double the A BIA.
Again, the parallelogram L BE M and the A EBC are upon the same base, BE; and between the same parallels, B.E and O M. Consequently, the parallelogram L BEM is double of the AEBC.
But the As ABI and EBC are equal to one another.
For the lines B I 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 <8 CBI and A BE are both right <s, and therefore equal to one another, if the < CBA be added to both, the sums are equal. That is to say, the _ A BI is equal to the < EBO.
Hence it appears, that the sides A B and BI, and the included 2 ABI in the one A, are equal respectively to the sides E B and B C, and the included
EBC in the other A. Therefore, the A ABI is equal to the A E B C. (Prop. IV.)
And since the doubles of equals are equal, it follows, that the square C BIH is equal to the parallelogram L BEM.
But if equals be added to equals, the sums are equal (Ax. II.)
Therefore the sum of the squares ACGF and BCHI is equal to the sum of the parallelograms ALMD and L BEM; that is, to the whole square A BED.
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 con
struction 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 Abe a right Z, the square
on the side which subtends the right is equal to the sum of the squares on the sides which
contain the right 2. (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).
Let A CB be a A, such, that, when squares are described
the three sides of it, the square described on the side A B is equal to the sum of the squares described on the sides A C and C B.
It has to be shown that the < AC B is a right Z.