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138. The diagonals of a parallelogram bisect each other.

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

the diagonals AC and B E cut each other at 0. We are to prove

A0 = 0 C, and BO= 0 E.

, =

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..A AO E= A BOC,

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

adj. s of the other).

..AO=OC,

and

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

Q. E. D.

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

А

E

B

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

diagonals AC and BE bisecting each other at 0.

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§ 108

.:.Δ ΑΟ Ε =Δ Α Ο Β, (having three sides of the one equal respectively to three sides of the other);

..LAOE-LAOB,

(being homologous É of equal A); ..LAO E and ZA OB are rt. [.

§ 25

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

Q. E. D.

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

В!

B

с

A

Α'

DI

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

A B = A' B', A D= A'D', and ZA= A'.
We are to prove that the S are equal.

Apply D A B C D to O 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 LA= 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';

.. the lines B C and B'C' coincide,

and C falls on B'C' or B'C' produced. In like manner D C and D' Care || 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 S ccincide, and are equal in all respects.

§ 66

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.

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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.

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Let S O be the straight line joining the middle points

of the non-parallel sides of the trapezoid A B C E.

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(being vertical 6 ).

.:. A FOE=AC OH,

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

adj. es of the other).

.. FE=CH,

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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