CASE III. If the sides opposite to BC be terminated in

the same point D,

W

B

the same method of proof is applicable,

but it is easier to reason thus:

Each of the s is double of ▲ BDC;

ABCD=□ DBCF.

I. 34.

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,

I. 35.

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

:.□ ABCD=□ EFGH.

Q. E. D.

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

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

Then must ▲ 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,

I. 35.

Now

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

EBCA,

I. 34.

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.

B

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

Then must A ABC= A 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.

Now A ABC is half of CG,

and A DEF is half of ☐ EH;
.. AABC= ADEF.

I. 36.

Ax. 7.

Q. E. D.

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.

the triangles CBD, ABE are equal.

Shew that

PROPOSITION XXXIX. THEOREM.

Equal triangles upon the same base, and upon the same side of it, are between the same parallels.

B

Let the equal As ABC, DBC be on the same base BC, and on the same side of it.

Join AD.

Then must AD be || to BC.

For if not, through A draw AO || to BC, so as to meet BD, or BD produced, in O, and join OC.

As ABC, OBC are on the same base and between

Then the same ||s,

the less-the greater, which is impossible;

.. AO is not || to BC.

I. 37

Hyp.

In the same way it may be shewn that no other line passing through A but AD is | to BC;

.. AD is | to BC.

Q. E. D.

Ex. 1. AD is parallel to BC; AC, BD meet in E; BC is produced to P so that the triangle PEB is equal to the triangle ABC: shew that PD is parallel to AC.

Ex. 2. If of the four triangles into which the diagonals divide a quadrilateral, two opposite ones are equal, the quadrilateral has two opposite sides parallel.