PROPOSITION XXXIV. THEOREM.

The opposite sides and angles of a parallelogram are equal to one another, and the diagonal bisects it.

B

D

Let ABDC be a , and BC a diagonal of the .
Then must AB-DC and AC=DB,

and

and

▲ BAC= 4 CDB, and ▲ ABD= 2 ACD
▲ ABC= ▲ DCB.

For AB is | to CD, and BC meets them, .. ABC alternate 4 DCB;

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and AC is to BD, and BC meets them,

.. ACB alternate

DBC.

I. 29.

Then in As ABC, DCB,

:: ▲ ABC= 2 DCB, and ▲ ACB= ▲ DBC,

and BC is common, a side adjacent to the equals in each ; .. AB=DC, and AC=DB, and ▲ BAC=▲ CDB,

Also

and A ABC= ▲ DCB.

ABC= 4 DCB, and 4 DBC= 4 ACB,

.. 4s ABC, DBC together= 4 s DCB, ACB together,

I. B.

that is,

L ABD= L ACD.

Q. E. D.

Ex. 1. Shew that the diagonals of a parallelogram bisect each other.

Ex. 2. Shew that the diagonals of a rectangle are equal.

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

Then must ☐ ABCD=□ EBCF.

CASE I. If AD, EF have no point common to both,

EBCF with ▲ EAB=figure ABCF;

ABCD with ▲ FDC=□ EBCF with ▲ EAB;

.. ABCD=EBCF.

CASE II. If the sides AD, EF overlap one another,

F

the same method of proof applies.

CASE III. If the sides opposite to BC be terminated in the same point D,

B

the same method of proof is applicable,

but it is easier to reason thus:

Each of thes is double of a BDC;

:.□ ABCD=□ DBCF.

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Q. E. D.

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

Let the Os ABCD, EFGH be on equal bases BC, FG, and between the same s AH, BG.

Then must ☐ ABCD=□ EFGH.

...they are on the same base BC and between the same ||s ;

and □ EBCH=□EFGH,

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they are on the same base EH and between the same ||s;

:. ABCD=□ EFGH.

Q. E. D.

PROPOSITION XXXVII. THEOREM.

Triangles upon the same base, and between the same parallels, are equal to one another.

D

Let As ABC, DBC be on the same base BC and between the same s AD, BC.

Then must A ABC= ^ DBC.

From B draw BE || to CA to meet DA produced in E.
From C draw CF || to BD to meet AD produced in F.

Then EBCA and FCBD are parallelograms,

and

EBCA=□ FCBD,

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Now

they are on the same base and between the same ||s.
A ABC is half of

EBCA,

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Ex. 1. If P be a point in a side AB of a parallelogram ABCD, and PC, PD be joined, the triangles PAD, PBC are together equal to the triangle PDC.

Ex. 2. If A, B be points in one, and C, D points in another of two parallel straight lines, and the lines AD, BC intersect in E, then the triangles AEC, BED are equal.

PROPOSITION XXXVIII. THEOREM.

Triangles upon equal bases, and between the same parallels, are equal to one another.

Let As ABC, DEF be on equal bases, BC, EF, and between the same ||s BF, AD.

Then must A ABC= ▲ DEF.

From B draw BG || to CA to meet DA produced in G. From F draw FH || to ED to meet AD produced in H. Then CG and EH are parallelograms, and they are equal, they are on equal bases BC, EF, and between the same Is BF, GH. I. 36.

Ex. 1. Shew that a straight line, drawn from the vertex of a triangle to bisect the base, divides the triangle into two equal parts.

Ex. 2. In the equal sides AB, AC of an isosceles triangle ABC points D, E are taken such that BD=AE. Shew that the triangles CBD, ABE are equal.