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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 Abe upon 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 ACG F and BIH C.
We have to prove
that the square ADEB A
is equal to the sum of the other two squares ACGF and BIH C.
From the point C 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 F B; C and D by the straight line C D; 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 BCH I is double of the A ABI; 6, that the parallelogram LBE M 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 ACG F is double of the A FAB.
Since G C A is an % of a square, it is a right 2: the _ AOB is also a right 2. Consequently, we have two straight lines, G C and BC, meeting a third, A C, at the same point C, and on opposite sides, and making the two adjacent Zs 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 FA.
Therefore the square ACGF is double the A FAB.
2. The parallelogram DMLA is double the A DAC; for CM 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 GC B is parallel to F A.
Therefore the parallelogram AD M L is double the ADAC.
3. The AFA B is equal to the ADAC.
For the lines FA and AC 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 Z8 FAC and D A B is an < of a square, they are right <s, and therefore equal to one another. Add to each of them the < CAB; and the sum of the Z8 FAC and C A B (i.e., the <F A B) will be equal to the sum of the Zs D A B and C AB (i.e., the < DA C).
Hence the two sides F A and A B and the < FAB in the A FAB, 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 F A B) 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 L BEM.
The proof of this part of the demonstration is of the same kind as that of the other.
The _ HCB 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 <s 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 B I.
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 BEM and the A 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 A EBC.
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 <s CBI and A BE are both right Zs, and therefore equal to one another, if the < CB A be added to both, the sums are equal. That is to say, the < A BI is equal to the < EBC.
Hence it appears, that the sides A B and BI, and the included Z A B I 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 O 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 AL MD and LBEM; 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.)
equal to each other. (Ax. I.)
square on the side which subtends the right _ is equal to the sum of the squares on the sides which
contain the right Z. (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 OB 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 < ACB is a right Z.