Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

PROPOSITION XXXIII. THEOREM.

The straight lines which join the extremities of two equal and parallel straight lines towards the same parts, are themselves equal and parallel.

Let AB, CD be = and || st. lines, joined towards the same parts by AC, BD.

Then AC, BD shall be = and II.

[blocks in formation]

Constr.

Join BC. Dem. Then, :: AB is 11 CD, (hyp.) and BC meets them, .: LABC= alt. ZBCD;

(i. 29) and :: AB = CD, (hyp.) and BC is common to the two As ABC, DCB, and LABC = _ BCD;

.. base AC = base BD,

(i. 4) and A ABC=ABCD, and the other <s= the other LS, each to each, to which the = sides are opp. ;

.. LACB = LCBD. And :: BC meets AC, BD, and makes the alt. Ls

ACB, CBD equal to one another; .. AC is || BD; (i. 27) and AC was shown = BD. Therefore, straight lines which, &c.

Q. E. D

PROPOSITION XXXIV. THEOREM.

The opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them, that is, divides them into two equal parts.

a

Let ACDB be a o of which BC is a diameter.
Then must AB =

CD; AC BD; and A ABC = ABCD.

B

(i. 29)

Dem. .: AB is 11 CD, and BC meets them, .. LABC= _BCD.

(i. 29) and :: AC is || BD, and BC meets them,

.. LACB= L CBD. Hence in As ABC, CBD, : the two 2. s ABC, BCA in the one, = BCD, CBD in the other, each to each; ; and one side BC, which is adj. to their = Ls is common ; .. their other sides are =, each to each, and the third

L of the one = the third L of the other; (i. 26) namely AB=CD, AC=BD, and _BAC= _BDC. And :: LABC= _ BCD and % CBD=LACB, .. whole Z.ABD = whole L ACD; (ax. 2) =

( and <BAC has been shown = LBDC. .. the opp. sides and Ls of a are equal to one another. Also, the diameter bisects it.

For since AB=CD, BC is common, and LABC has been shown =_ BCD;

= .. A ABC = A BCD; (i. 4) and the diameter bisects it.

Q. E. D.

[ocr errors]

[Avoid two common errors : (1) when applying prop. 26 do not say therefore the triangles are equal in all respects;' (2) do not stop short after proving that the opposite sides and angles are equal. It is necessary to complete the proof by applying prop. 4. This prop. is the last in the second section of the first Book, which treats of the theory of parallel lines.]

1. In what case will the diagonal bisect the angle of a parallelogram?

2. Straight lines which bisect two opposite angles of a parallelogram are parallel or coincident.

3. Draw lines through the angular points of a parallelogram which shall form another parallelogram double of the former.

4. The sides AB, AC of a triangle are bisected in D, E respectively, and BE, CD are produced till EF=EB, and GD=DC; show that the line GF passes through A.

5. From the angles C, B of a parallelogram ABCD, CE, BE are drawn parallel to the diagonals, which EA, ED intersect in H, K. Show that HK is half BC.

6. The diagonals of a rhombus bisect one another at right angles.

7. Through a given point between two straight lines which are not parallel draw a straight line which shall be bisected in that point.

8. Draw a line from a given point in a side of a parallelogram which shall bisect the parallelogram.

9. Bisect a parallelogram by a line drawn at right angles to one of the sides.

The diagonals AC, BD of a parallelogram inter

sect in 0, and the parallelograms AOBP, DOCQ are formed. Show that POQ is a straight line.

II. The diagonals of a parallelogram are equal : find its angles.

12. Show that a given straight line may be thus trisected; on it as diagonal describe any parallelogram: draw lines from the middle points of two opposite sides to two opposite angles : these lines trisect the given line.

PROPOSITION XXXV. THEOREM.

Parallelograms on the same base, and between the same parallels, are equal to one another.

Let the OS ABCD, EBCF be on the same base BC, and between the same ||s AF, BC.

Then shall ABCD -- EBCF.

[ocr errors]
[blocks in formation]

Dem. If the sides AD, DF of OS ABCD, DBCF, opposite BC, be terminated at the same pt. D; then, since each ois double A DBC; (i. 34)

ABCD =ODBCF. (ax. 6) But if AD, EF be not terminated at the same pt.; Then, : ABCD is o .. AD = BC; (i. 34)

z for a similar reason EF=BC; .. AD=EF ;

(ax. 1) and DE is common; .. the whole (or the remainder)

AE=the whole (or the remainder) DF, (ax. 2 or 3) :, in As EAB, FDC, FD=AE, DC=AB, (i. 34) and ext. LFDC=int. _EAD; (i. 29) .. AFDCA EAB.

(i. 4)

a

« ΠροηγούμενηΣυνέχεια »