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PROPOSITION XLII. THEOREM. 138. The diagonals of a parallelogram bisect each other.

A

Let the figure A B C E be a parallelogram, and let

the diagonals A C and BE cut each other at 0.
We are to prove A0 = 0 C, and B 0= 0 E.
In the A A O E and BOC
A E= BC,

§ 134
(being opposite sides of a 0),
ZOA E=LOCB,

§ 68 (being alt.-int. £), ZOEA = LOBC;

§ 68 .:. A A O E=A BOC,

§ 107 (having a side and two adj. of the one equal respectively to a side and two

adj. És of the other).
.: A(=0C,

BO= 0 E.
(being homologous sides of equal A ).

and

Q. E. D.

PROPOSITION XLIII. THEOREM.

139. The diagonals of a rhombus bisect each other at right angles.

Let the figure A B C E be a rhombus, having the

diagonals AC and BE bisecting each other at 0.

$ 128

We are to prove. Z AO E and L A OB rt. £.
In the A AO E and A O B,

A E = A B,
(being sides of a rhombus) ;

O E= 0 B,
(the diagonals of a o bisect cach other) ;

A (= A 0,

$ 138

Iden.

..A AO E= A A OB,

$ 108 (having three sides of the one equal respectively to three sides of the other);

..ZA O E= L A OB,

(being homologous é of equal A);
.. LAO E and ZA O B are rt. £.

$ 25

(When one straight line meets another straight line so as to make the adj. És equal, each Lis a rt. 2).

Q. E. D.

PROPOSITION XLIV. THEOREM. 140. Two parallelograms, haring two sides and the included angle of the one equal respectirely to two sides and the included angle of the other, are equal in all respects.

B

B

In the parallelograms A B C D and A' B' C'D', let

AB= A' B', A D= A'D', and LA = A'.
We are to prove that the are equal.

Apply D A B C D to A' B'C' D', so that A D will fall on and coincide with A' D'.

Then A B will fall on A' B',

(for ZA= L A', by hyp.),
and the point B will fall on B'.

(for A B = A' B', by hyp.). Now, BC and B'C' are both || to A' D' and are drawn through point B';

is the lines B C and B'Ccoincide, § 66

and C falls on B'C' or B'C' produced. In like manner DC and D' Care Il to A' B' and are drawn through the point D'. . DC and D' C' coincide;

§ 66 ... the point C falls on D' C', or D' C' produced ;

.:. C falls on both B' C and D' C";
::: C must fall on a point common to both, namely, C'.
.:. the two 5 ccincide, and are equal in all respects.

Q. E, D. 141. COROLLARY. Two rectangles having the same base and altitude are equal; for they may be applied to each other and will coincide.

PROPOSITION XLV. THEOREM.

142. The straight line which connects the middle points of the non-parallel sides of a trapezoid is parallel to the parallel sides, and is equal to half their sum.

Let S O be the straight line joining the middle points

of the non-parallel sides of the trapezoid A BC E.

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.:. A FOE=AC OH,

§ 107 (having a side and two adj. of the one equal respectively to a side and two .. FE=CH,

adj. s of the other).

and

OF= 0 H,
(being homologous sides of equal ).

Now

$ 135

FH= AB,
(!l lines comprehended between Il lines are equal);

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S O is also || to BC,
(a straight line II to one of two II lines is II to the other also).

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Substitute for A F and BH their equals, A E FE and BC+CH,

and add, observing that CH= FE;

then

2 80= A E + BC.

..SO= }(A E + BC).

Q. E. D.

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