And, because AB is equal to DC (by Hyp.), AC com. mon to each of the triangles ABC, ADC, and the angle DCA equal to the angle CAB, the fide AD will also be equal to the fide BC, and the angle DAC to the angle ACB (Prop. 4.)

Since, therefore, the right line AC interfects the two right lines AD, BC, and makes the alternate angles equal to each other, thofe lines will be parallel (Prop. 23.), ...

But the line AD has been proved to be equal to the line BC; confequently they are both equal and parallel. Q.E.D.

PROP. XXX. THEOREM.

The oppofite fides and angles of any paral lelogram are equal to each other, and the diagonal divides it into two equal parts,

Let ABCD be a parallelogram, whofe diagonal is AC; then will its oppofite fides and angles be equal to each other, and the diagonal AC will divide it into two equal parts.

For, fince the fide AD is parallel to the fide BC (Def, 22.), and the right line ac interfeets them, the angle DAC will be equal to the alternate angle ACB (Prop. 24.)

And, because the fide DC is parallel to the fide AB (Def. 22.), and AC intersects them, the angle PCA will be equal to the alternate angle CAB (Prop. 24.)

Since, therefore, the two angles DAC, DCA, are equal to the two angles ACB, CAB, each to each, the remaining

angle ADC will also be equal to the remaining angle ABC (Prop, 29. Cor.) and the whole angle DAB to the whole angle DCB.

But, the triangles CDA, ABC, being mutually equiangular, and having AC common, the fide DC will also be equal to the fide AB, and the fide AD to the fide BC, and the two triangles will be equal in all respects (Prop 21.) Q.E.D.

PROP. XXXI. THEOREM.

Parallelograms, and triangles, ftanding upon the fame bafe, and between the fame parallels, are equal to each other.

5. Let AE, BD be two parallelograms standing upon the fame base AB, and between the fame parallels AB, DE; then will the parallelogram AE be equal to the parallelogram BD.

For, fince AD is parallel to BC (Def: 22.), and DE intersects them, the outward angle ECB will be equal to the inward oppofite angle FDA (Prop. 25.)

And, because AF is parallel to BE (Def. 22.), and DE interfects them, the outward angle AFD will be equal to the inward oppofite angle BEC (Prop. 25.)

> Since, therefore, the angle ECB is equal to the angle FDA, and the angle AFD to the angle BEC, the remaining

angle CBE will be equal to the remaining angle DAY (Prop. 29. Cor. 1.)

But the fide AD is alfo equal to the fide BC (Prop. 31.); R confequently, fince the triangles ADE, BCE are mutually equiangular, and have two correfponding fides equal to each other, they will be equal in all refpects (Prop. 21.)

If, therefore, from the whole figure ABED, there be taken the triangle BCE, there will remain the parallelogram BP; and if, from the fame figure, there be taken the triangle ADF, there will remain the parallelogram AE.

But if equal things be taken from the fame thing, the remainders will be equal; confequently, the parallelogram AE is equal to the parallelogram BD.

Again, let ABC, ABF be two triangles, ftanding upon the fame base AB, and between the fame parallels, AB, CF; then will the triangle ABC be equal to the triangle

ABF.

For produce CF, both ways, to D and E, and draw AD parallel to BC, and BE to AF (Prop. 28.)

Then, fince BD, AE, are two parallelograms, standing upon the fame bafe AB, and between the fame parallels AB, DE, they are equal to each other (Prop. 32.)

And, because the diagonals AC, BF bifect them (Prop. 31.), the triangle ABC will alfo be equal to the triangle. ABF. Q.E.D.

PROP. XXXII,

THEOREM.

If a parallelogram and a triangle stand. upon the fame bafe, and between the fame parallels, the parallelogram will be double the triangle.

Let the parallelogram AC and the triangle ALB stand upon the fame base AB, and between the fame parallels AB, DẸ; then will the parallelogram AC be double the triangle AEB.

For join the points B, D; then will the parallelogram AC be double the triangle ADB, because the diagonal DB divides it into two equal parts (Prop. 31.)

But the triangle ADB is equal to the triangle A EB, be cause they ftand upon the fame base AB, and between the fame parallels AB, DE (Prop. 32.) ; whence the parallelogram AC is alfo double the triangle AEB. Q. E. D.

COROLL. If the bafe of the parallelogram be half that of the triangle, or the base of the triangle be double that of the parallelogram, the two figures will be equal to each other.

PROP.

PROP. XXXIII. PROBLEM.

To make a parallelogram that shall have its oppofite fides equal to two given right lines, and one of its angles equal to a given rectilineal angle,

Let AB and C be two given right lines, and D a given rectilineal angle; it is required to make a parallelogram that shall have its oppofite fides equal to AB and c, and one of its angles equal to D.

At the point A, in the line AB, make the angle BAF equal to the angle D (Prop. 20.) and the fide AF equal to c (Prop. S.)

Alfo, make FE parallel and equal to AB (Prop. 28 and S.), and join BE; then will AE be the parallelogram required.

For, fince FE is parallel and equal to AB (by Conft.), BE will be parallel and equal to AF (Prop. 30.); whence the figure A is a parallelogram.

And, because AF is equal to c (by Conft.) BE will alfo be equal to C; and the angle BAF was made equal to the angle D.

The oppofite fides of the parallelogram AE are, therefore, equal to the two given lines AB and C; and one of its angles is equal to the given angle D, as was to be done.