PRO P. XXVIII. THEOREM.

If one fide of a triangle be produced, the outward angle will be equal to the two inward oppofite angles, taken together; and the three angles of every triangle, taken together, are equal to two right angles.

N

Let ABC be a triangle, having one of its fides AB produced to D; then will the outward angle CBD be equal to the two inward oppofite angles BCA, CAB, taken together; and the three angles BCA, CAB and ABC, taken to gether, are equal to two right angles.

For through the point B, draw the right line BE parallel to AC (Prop. 28.).

Then, because BE is parallel to AC, and CB intersects them, the angle CBE will be equal to the alternate angle BCA (Prop. 24.)

And because BE is parallel to AC, and AD interfects them, the outward angle EBD will be equal to the inward angle CAB (Prop. 25.)

But the angles CBE, EBD are equal to the whole angle CBD; therefore the outward angle CBD is equal to the two inward oppofite angles BCA, CAB taken together.

And if, to thefe equals, there be added the angle ABC, the angles CBD, ABC, taken together, will be equal to the three angles BCA, CAB and ABC, taken together.

But the angles CBD, ABC, taken together, are equal to two right angles (Prop. 13.); confequently the three angles BCA, CAB and ABC, taken together, are alfo equal to two right angles.

COROLL. 1. If two angles of one triangle, be equal to two angles of another, each to each, the remaining angles will also be equal.

COROLL. 2. Any quadrilateral may be divided into two triangles; therefore all the four angles of fuch a figure, taken together, are equal to four right angles.

.PRO P. XXIX.

THEOREM.

Right lines joining the correfponding extremes of two equal and parallel right lines are themselves equal and parallel.

Let AB, DC be two equal and parallel right lines; then will the right lines AD, BC, which join the correfponding extremes of those lines, be also equal and parallel.

For draw the diagonal, or right line AC :

Then, because AB is parallel to DC, and AC intersects them, the angle DCA will be equal to the alternate angle CAB (Prop. 24.)

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 intersects the two right lines AD, BC, and makes the alternate angles equal to each other, those 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.

PRO P. XXX. THEOREM.

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

Let ABCD be a parallelogram, whose 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 intersects 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 interfects them, the angle DCA 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.

W

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

gram 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 DA, and the angle AFD to the angle BEC, the remaining angle

D 4

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

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

If, therefore, from the whole figure ABED, there be taken the triangle BCE, there will remain the parallelogram BD; 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, standing 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 cr, 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 base 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.. $1.), the triangle ABC will also be equal to the triangle ABF. Q.E.D.