By this Problem 'tis eafy to conceive how to make a Square equal to any given Parallelogram. (14. e. 6.)

For if BP be the Length, and P C be the Breadth of the given Parallelogram, then will AP be the Side of the Square, equal in Area to that Parallelogram.

Three Right-lines being given, to find a fourth Proportional Line.

Suppofe the three Lines

(12. e. 6.)

Upon the longeft Line AB fet off the next longeft Line A D; viz. make D BAB AD; then upon the Poin D fet the other Line

DC at an Angle, either right or oblique, and draw the Right-line AC, continuing it a fufficient Length; make B Fparallel to DC, and it will be the fourth Proportional requir'd; that is, AD: DC:: AB: BF.

THEOREM XIV.

If any Angle of a plain Triangle be bifected (viz. divided into two equal Angles) with a Right-line, viz. as C A is fuppos'd to do the Angle B CD) it will cut the oppofite Side (viz. B D) in Proportion to the other two Sides of the Triangle. (3. e. 6.) i. e. BA: BC::AD: CD.

Demonftration.

Produce the Side D C, until CZ CB: join the Points Z B with a Right line, and draw the Line FC parallel to BD; whence the <2=CBZ per Theorem 6. and <Z+CBZ, or 2 CBZS BCD per Theorem 5; or, dividing both Sides of the Equation by 2, CBZ = 1 BCD. But ¦ <

BCD=ACB

F

B

ACD by the Hypothefis, therefore ACB CBZ per Axiom 5: Whence AC is parallel to B Z per Theorem 3, and the Triangles BDZ, ADC, and FCZ are fimilar by the fecond Figure to Theorem 12. confequently BA (= FC): BC (ZC):: AD: CD. Q. E. D.

THE O

THEOREM XV.

If two Right-lines (howfoever drawn) within a Circle do cut each other, the Rectangle made of the Segments (or Parts) of the one Line, will be equal to the Rectangle made of the Segments (or Parts) of the other Line. (35. e. 3.)

That is, if two Lines (as AB and CD) any Point, as at x, then will Ax x B x

Demonftration.

Join the Points AC and BD with Right-lines, then will the ▲ Cx A be like to AB x D: For B< C and < A = < D. by Corollary to Theorem 9. and AxC Bx D. by Theorem 2. Therefore it will be Ax:

do cut each other in D x × Cx.

C

B

D

Dx:: Cx: Bx. by Theorem 13. Confequently Ax x B x = DXXGx. Q. E. D.

THEOREM XVI.

If two Right-lines are fo drawn within a Circle, as, being continued, they will meet in a Point out of the Circle's Periphery, the Rectangle made of the one whole Line, and its Part out of the Circle, will be equal to the Rectangle of the other whole Line, and its Part out of the Circle. (36, 37. c. 3.)

That is, if the Lines AC and D B be continued unto the Point Z; then will AZ x CZ =DZXBZ.

Demonftration.

Draw the Lines AB and CD, then will A CZ D be like to the

ABZ A; for

A=D, and

angles, confequently, ABZ

Z is common to both Tri< DC Z, by Theorem 4. therefore AZ BZDZ: CZ. Ergo, AZ x C Z = : DZXBZ.

THEOREM XVII.

If from any Angle of a plain Triangle infcrib'd in a Circle there be let fall a Perpendicular upon the oppofite Side, as D P;

As that Perpendicular is in Proportion to one of the Sides including the Angle, fo is the other Side including the Angle to the Di̟ameter of the Circle.

If any Quadrangle (that is, a Trapezium) be infcrib'd within a Circle, the two oppofite Angles, taken together, are equal to two Right-angles, viz. 18. (22. e. 3.)

That is, in the Quadrangle A B C D the < A+C= 180°. And the B+D=180°..

BCA + B AC = 180°. by

Theorem 4. and the BD A+

<

BDC

B

<ADC. Therefore the ABC ADC 180°. and by the fame Way of arguing it may be prov'd, that the < BAD+BCD 180. Q. E. D.

THEOREM XIX.

If in any Quadrangle infcrib'd within a Circle there be drawn two Diagonals, as AC and BD, the Rectangle made of the two Diagonals will be equal to both the Rectangles made of the oppofite Sides of the Quadrangle.

That is, ACX B D = ABX CD + ADX BC.

Demon

Demonstration.

Make the Arch DG

Arch BC, and from the Points A, G draw the Line Af, and it will form the A AfD, like to the AABC: For the ƒ A D = BAC, because the Arches DG and BC are equal.

Again, the fDA=< BC A, becaufe they both ftand upon the Arch AB: Confequently, the AfD ABC,

A

D

B

by Theorem 4. Therefore it will be AC: BC::AD: Df, by

Theorem 13. Ergo

BCXAD

AC

=Df.

Again, the A B Af and A ACD are alike: For A Bf ACD, and B Af CAD, becaufe the fAD =<BAC, and the C Af is common to both Triangles. Confequently, the Aƒ BADC. Therefore AC: CD ::AB: Bf, by Theorem 13. Ergo

CDX AB

AC

= Bf. But DfBf=BD. Confequently, BCXAD+CD XAB

=BDX AC. Q. E. D.

THEOREM XX.

All Parallelograms (whether Right or Oblique-angled) that fland upon the fame Bafe, or upon equal Bases, and betwixt the fame Parallels, are equal to one another. (35. & 36. e. 1.)

That is, A B C D = — a b C D:

Demonstration.

A

B

a

b

Becaufe ABCD a b, by Suppofition, therefore AaBb; for Ba is common to both. And becaufe ACB D, and the AS B, therefore the AAC a▲ B Db: And if from both Triangles there be taken the A B x a common to both,

there will remain the Trapezium AB x G c abx D, per Axiom 5.

D

But

But the Trapezium A B x C + AC x D = □ ABC D. and the Trapezium ab x D + A CxD=ab CD. confequently, ABCD a b C D. Q. E. D,

Corollary.

Hence it will be eafy to conceive, that all Triangles which ftand upon the fame Bafe, or upon equal Bafes, and between the fame Parallels, (viz. having the fame Height) are equal one to another. (37 & 38. e. 1.)

For all Triangles are the Halfs of their circumfcribing Parallelograms; and therefore, if the Wholes be equal, their Halfs will alfo be equal.

THEOREM XXI.

Parallelograms (and confequently Triangles) which have the fame Height, have the fame Proportion one to another as their Bases have, (I. e. 6.)

Like Triangles are in a duplicate Ratio to that of their homologous Sides. (19. e. 6.)

That is, the Area's of like Triangles are in Proportion one to a nother as are the Squares of their like Sides.