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PROPOSITION. VIII.

THEOREM.
If a Perpendicular be drawn in a Right-angled

Triangle, from the Right Angle to the Base,
tben ibe Triangle on each side of the Perpendi-
cular are similar, both to the Wbole, and also to
one another.

1

LETA BC be a Right-angled Triangle, whose

Right Angle is BAC; and let the Perpendicular A D, be drawn from the Point A to the Base B C. I fay the Triangles ABD, ADC, are similar to one another, and to the whole Triangle ABC.

For, because the Angle B A C is equal to the An. gle A D B, for each of them is a Right Angle ; and the Angle at B is common to the two Triangles A B C, A BD; the remaining Angle A C B shall be * equal to the remaining Angle B AD. Therefore * Cor. 32. 16 the Triangle A B C is equiangular to the Triangle ABD; and so, as + BC, which fubtends the Right + 4 of this. Angle of the Triangle A B C, is to B A, subtending the Right Angle of the Triangle ABD, fo is AB, fubtending the Angle C of the Triangle A BC, to DB, subtending an Angle equal to the Angle C, viz. the Angle B AD, of the Triangle ABD; and so, moreover, is AC to AD, subtending the Angle B, which is common to the two Triangles. Therefore the Triangle A B C is fequiangular to the Triangle 1 Def. in ABD; and the sides about the equal Angles are this. proportional. Wherefore the Triangle A B C is I also similar to the Triangle ABD. By the same Way we demonftrate, that the Triangle ADC is also fimilar to the Triangle A B C. Wherefore each of the Triangles A B D, A DC, is similar to the whole Triangle.

I say, the said Triargles are also similar to one rother,

For, because the Right Angle B D A is equal to the Right Angle ADC, and the Angle BAD has been proved equal to the Angle C; it follows, that the remaining Angle at B * Ihall be equal to the remaining * Cor. 32. 1.

Angle

Angle D AC. And so the Triangle A B D is equi† 4 of sbiso angular to the Triangle A DC. Wherefore as +

BD, subtending the Angle B A D of the Triangle ABD, is to D A, subrending the Angle at C of the Triangle ADC, which is equal to the Angle B AD; fo is A D, fubtending the Angle B of the Triangle ABD, to D C, subtending the Angle D A C, equal to the Angle B. And, moreover, so is B A to AC, fubtending the Right Angles at D; and, consequently, the Triangle A B D is similar to the Triangle A DC. Wherefore, if a Perpendicular be drawn in a Rightangled Triangle, from the Right Angle to the Base, then the Triangles on each side of the Perpendicular are similar, both to the Whole, and also to any another ; which was to be demonstrated.

Coroll. From hence it is manifest, that the Perpendi

cular drawn in a Right-angled Triangle from the Right Angle to the Base, is a mean Proportional between the Segments of the Base. Moreover, either of the sides containing a Right Angle, is a mean Proportional between the whole Base, and the Segment thereof, which is next to the Side.

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PROPOSITION IX.

PROBLEM
To cut off any Part required from a given Right

Line.
LET AB be a Right Line

given, from which must be cut off any required Part; suppose a third. Draw any Right Line AC from the Point A, making an Angle at Pleasure with the Line A B. Affume

any Part Din the Line A C; make * DE, EC, each # 31. 1.

equal to AD; join BC, and draw + D F through D, parallel to B Č.

Then, because F D is drawn Parallel to the Side | 2 of tbis, BC of the Triangle A B C, it Thall be, I as C D is to

DA, fo is BF to F A. But CD is double to D A. Therefore B F shall be double to FA; and so B A is triple to AF. Wherefore, there is cut off A F, a third Part required of the given Right Line AB; which was to be done.

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3. I.

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PROPOSITION X.

PROBLEM.
To divide a given undivided Right Line, as an-

other given Right Line is divided.
LE ET A B be a given undivided Right Line, and

A C a divided Line. It is required to divide A B,
as A C is divided,

Let A C be divided in the Points D and E, and so placed, as to contain any Angle with A B. Join the Points C and B ; through D and E let DF, ÉG, be drawn * parallel to B Č; and thro' D draw D HK,* 31.1. parallel to A B.

Then FH, HB, are each of them Parallelograms; and so D H is t equal to FG, and H K to G B. And +34. 1. because HE is drawn parallel to the Side KC of the Triangle D KC, it shall be I as CE is to E D, so is 1 2 of tbis. KH to HD. But K H is equal to B G, and H D to GF. Therefore, as CE is to ED, so is B G to GF. Again, because F D is drawn parallel to the Side E G of the Triangle A GE, as E D is to D A, fo fhall + GF be to F A. But it has been proved, that CE is to ED, as B G is to GF. Therefore, as CE is to E D, fo is B G to GF; and as E D is có DA, so is G F to FA. Wherefore, the given undivided Line A B is divided as the given Line A Cis; which was to be done.

PROPOSITION XI.

PROBLEM.
Two Right Lines being given, to find a third

Proportional to thim.
LET AB, AC, be two given Right Lines, so

placed, as to make any Angle with each other.
It is required to find a third Proportional to AB, AC.

Produce AB, AC, to the Points D and E, make
BD equal to AC; join the Points B, C; and draw
I the Right Line D E thro’D, parallel to BC.
M

Then,

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Then, because B C is drawn parallel to the Side | 2 of this. DE of che Triangle ADE, it fhall be, I as A'Bis

to BD, so is AC to CE. But B D is equal to AC Hence, as A B is to AC, so is A C to CE. Therefore, a third Proportional CE is found to two given Right Lines A B, A C; which was to be done.

PROPOSITION XII.

PROBLEM.
Three Right Lines being given, to find a fourth

Proportional to ibem.
L!
ET A, B, C, be three Right Lines given. It is

required to find a fourth Proportional to them.
Let D E and D F be two Right Lines, making any
Angle EDF with each other. Now make D G equal
to A, G E equal to B, D H equal to C; and draw the
Line GH, as also + E F thro' E, parallel to G H.
. Then, because G H is drawn parallel to EF, the
Side of the Triangle D EF, it fhall be, as D G is to
GE, fo is D H to HF. But D G is equal to A,
GE to B, and DH to C. Consequently, as A is to
B, fo is C to HF. Therefore, the Right Line H F, a
fourth Proportional to the three given Right Lines A, B,
, is found ; which was to be done.

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PROPOSITION XIII.

PROBLEM
To find-a mean Proportional between two given

Right Lines.
LE
ET the two given Right Lines be A B, BC. It

is required to find a mean Proportional between then. Place AB, BC, in a direct Line ; and on the Whole AC describe the Semicircle ADC, and

I draw B D at Right Angles to AC from the Point 111. I.

B; and let AD, D C, be joined.

Then, because the Angle AD.C, in a Semicircle, +31. 3. is + a Right Angle ; and since the Perpendicular DB is drawn from the Right Angle to the Bale; therefore

DB

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DB is f a mean Proportional between the Segments of 1 Cor. 8. of the Base A B, B C." Wherefore, a mean Proportional ibis. between the two given Lines A B, BC, is found; which was to be done.

PROPOSITION XIV.

THEOREM.
Equal Parallelograms, having one Angle of the

one equal to one Angle of the other, bave the
Sides about the equal Angles reciprocal ; and
those Parallelograms that have one Angle of
the one equal to one Angle of the other, and the
Sides that are about the equal Angles recipro-

cal, are equal LET A B, B.C, be equal Parallelograms, having

B B E, be in one strait Line ; then also I will the Sides (14. t. FB, B G, be in one strait Line. I say, the sides of the Parallelograms AB, BC, that are about the equal Angles, are reciprocal; that is, as A D is to B E, lo is G B to BF.

For, let the Parallelogram F E be compleated.

Then, because the Parallelograms A B is equal to the Parallelogram B C, and F E is some other Parallelogram ; it shall be, as A B is to F E, so is + B C +7.5. to FE; but as A B is to F E, so is IBD to BE; and as B C is to F E, so is G B to B F. Therefore 11 of sbit. as D B is to BE, so is G B to B F. Wherefore the Sides of the Parallelograms A B, B C, that are about the equal Angles, are reciprocally proportional.

And, if the Sides that are about the equal Angles are reciprocally proportional, viz. if D B be to B E, as GB ir to BF; I say, the Parallelogram A B is equal to the Parallelogram BC.

For, fince D B is to B E, as G B is to BF; and D B to B E, as the Parallelogram A B I to the Parallelogram F E, and G B I to B F, as the Parallelogram B C to the Parallelogram FE; it shall be as AE is to FE, fo is B C F to FE. Therefore the Parallelogram A B is equal to the Parallelogram B C. · And so, equal Parallelograms, having one Angle of the

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