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the sides AB, BC are equal to DC, CB, each to each ; and the angle ABC was proved to be equal to DOB; therefore the triangles ABC, BCD are equal.
(I. 4) That is, the diameter BO bisects the parallelogram ACDB. Wherefore, the opposite sides, &c.
Q. E. D.
1. Show how we might superpose one of the triangles in the figure of XXXIV. upon the other.
2. Show how an extended enunciation of Prop. XXVI. would enable us to shorten the demonstration of XXXIV.
3. One angle of a parallelogram is half a right angle. Find the others.
4. One angle of a parallelogram is one-third of two right angles. Find the remaining angles.
5. If the parallelogram just referred to be equilateral, prove that one of its diagonals will divide it into two equilateral triangles.
6. If one angle of an equilateral quadrilateral be right, show that the figure is a square.
(Questions 2–6 refer to Diagram II., 7-9 to Diagram IV.) 1. State a fourth condition, derived from XXXIII., upon which two straight lines will be parallel.
2. PQ, UV are equal and parallel ; name two other lines which must also be equal and parallel.
3. Given ABCD a parallelogram, write out the consequences arising by XXXIV.
4. Write out similar consequences with regard to the parallelogram EHDX.
5. Given APEY, PQEY, both parallelograms ; prove AQ bisected in P. 6. Also prove that the triangles APY, PQE are equal.
7. AB is equal to A0 and parallel to 0C; prove that BC is equal and parallel to 0 A.
8. AF, in the same, is also equal to 0 A and parallel to OE; prove A FEO a parallelogram.
9. Given the results of 7 and foregoing, prove BF equal and parallel to CE.
SECTION IX.-PROPOSITIONS XXXV.-XLII. Parallelism in relation to Equivalent Parallelograms
and Triangles. 82. THE PECULIARITIES OF PROPOSITION XXXV. The manner in which the next proposition is proved deserves notice on three points.
Firstly, we have Axioms 2 and 3 used in a modified form, which we may express thus (see Art. 19 also) :
2. If the same thing be added to equals, the wholes are equal.
3. If the same be taken from equals, the remainders are equal. Secondly, we have Axiom 3 further modified so as to be equivalent to the following :
If two different parts of one whole be equal, and if one be taken away and restored again, and then the other be taken away, the remainder after the first operation will be equal to the remainder
after the second. Thirdly, the principal demonstration is made to apply, excepting in one line, to two figures at the same time ; yet not altogether in the same way. This occurs here for the first time with us; it often occurs in Geometry, but will not do so again in this book.
The two figures above referred to should be drawn by the learner together; and it should be carefully observed how each step of the argument applies to both.
EXERCISE XLIV. 1. Given AB, CD equal; enunciate an axiom
A B с D which proves AC equal to BD; and state in which of Euclid's axioms it is included.
2. Given AB and CD equal again ; enunciate an A CB D axiom proving AC equal to BD. Which of Euclid's axioms includes it?
3. If we say, in reference to the lines above given, “The whole, or the remainder, AC is equal to the whole, or the remainder, BD"; which of these words will refer to the upper line alone, which to the second alone, and which to both ?
4. In the annexed figure, from the trapezium ABCD take the triangle ACD; what is left ?
5. Having put ACD back again, take ABD away ; what is left ?
6. Given, in the same figure, that the triangles ABC, DBC are equal, apply a form of Axiom 3 to prove that the triangles ABD and ACD are also equal.
Note.—That part of Proposition XXXV. which precedes the first figure should be read with reference to all the figures. If the second part of the proposition were taken first, the order would be improved in respect of convenience.
83. PROPOSITION XXXV.-THEOREM. Parallelograms upon the same base and between the same parallels
are equal to one another. Let the parallelograms ABCD, EBCF be upon the same base
BO, and between the same parallels AF, BC.
First, let the sides which are opposite the base BC be ter
minated in the same point, so that E coincides with D, and the parallelograms become
ABCD, DBOF. Then it is plain that each parallelogram is double of the triangle BDC;
(I. 34) therefore ABCD is equal to DBCF.
Secondly, let the sides AD, EF, opposite the base BC, be
not terminated in the same point. Then, because ABCD is a parallelogram, therefore AD is equal to BC;
(I. 34) and, for a similar reason, EF is equal to BC; wherefore AD is equal to EF.
(Ax. 1) Therefore the whole, or the remainder, AE is equal to the whole, or the remainder, DF. Also, AB is equal to DC; hence, in the triangles, FDC, EAB, D E
(Ax. 2, or 3)
FD, DC are equal to EA, AB, each to each ; and the exterior angle FDC is equal to the opposite interior angle EAB;
(I. 29) therefore the triangle FDC is equal to the triangle EAB. (1.. 4) From the trapezium ABCF take the triangle FDC; the paral
lelogram ABCD remains : from the same trapezium take the triangle EAB* ; the paral
lelogram EBCF remains; and these remainders are equal;
(Ax. 3) that is, the parallelograms ABCD, EBCF are equal. Therefore, parallelograms upon the same base, &c.
Q. E. D. EXERCISE XLV. 1. By means of I. 29, find two angles in FDC which are equal to two in EAB, each to each.
What proposition shows that the remaining angles in those triangles are equal ?
2. Given that these remaining angles are equal, prove the triangles FDC, EAB are equal without using Axioms 2 or 3. (Apply I. 4 to the triangles.)
3. In the figure of XXXVI., if EB be equal and parallel to CH, what must be true of EH, BC according to XXXIII. ?
4. What would then follow with regard to the figure EBCH ?
5. If EBCH and EFGH be said to be upon the same base, which line will be meant ?
6. Prove these figures equal if they are parallelograms and BG is one straight line.
* For this operation the triangle FDC must be supposed to have been put back again.
Parallelograms upon equal bases and between the same parallels
are equal to one another. Let ABCD, EFGH be parallelograms upon equal bases BC,
FG, and between the same parallels AH, BG.
Because BC is equal to FG (Hyp.), and FG to EH;(I. 34) therefore BC is equal to EH.
(Ax. 1) Also, these lines are parallels, and joined towards the same parts by BE, CH. But straight lines which join the extremities of equal and
parallel straight lines towards the same parts, are themselves equal and parallel ;
(I. 33) therefore BE, CH are both equal and parallel; wherefore EBCH is a parallelogram.
(Def. A.) Then, since the parallelograms ABCD, EBCH are upon the
same base BC, and between the same parallels BC, AH; therefore ABCD is equal to EBCH.
(I. 35) For a similar reason, EFGH is equal to EBCH. Hence the parallelograms ABCD, EFGH are equal to one another.
(Ax. 1) Therefore parallelograms upon equal, &c.
Q. E. D. EXERCISE XLVI. 1. Prove XXXVI., by joining AF, DG instead of BE, CH.
2. Make the figure of XXXVI., and join EC; then name two triangles which are each half of EBCH.