PROPOSITION VIII.

THEOREM.

If a Perpendicular be drawn in a Right-angled Triangle, from the Right Angle to the Bafe, then the Triangle on each Side of the Perpendicular are fimilar, both to the Whole, and also to one another.

LET ABC be a Right-angled Triangle, whose Right Angle is BAC; and let the Perpendicular A D, be drawn from the Point A to the Bafe BC. I fay the Triangles ABD, ADC, are fimilar to one another, and to the whole Triangle A B C.

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For, because the Angle B A C is equal to the Angle A D B, for each of them is a Right Angle; and the Angle at B is common to the two Triangles ABC, ABD; the remaining Angle A C B shall be equal to the remaining Angle BAD. Therefore Cor. 32. 1% the Triangle ABC is equiangular to the Triangle ABD; and fo, as + BC, which fubtends the Right † 4 of this. Angle of the Triangle A B C, is to B A, fubtending the Right Angle of the Triangle ABD, fo is A B, fubtending the Angie C of the Triangle A B C, to DB, fubtending an Angle equal to the Angle C, viz. the Angle B AD, of the Triangle ABD; and fo, moreover, is A C to AD, fubtending the Angle B, which is common to the two Triangles. Therefore the Triangle ABC is equiangular to the Triangle 1 Def. 1. f ABD; and the Sides about the equal Angles are this. proportional. Wherefore the Triangle A B C is alfo fimilar to the Triangle A B D. By the fame Way we demonftrate, that the Triangle ADC is alfo fimilar to the Triangle A B C. Wherefore each of the Triangles A B D, A D C, is fimilar to the whole Triangle.

I fay, the faid Triangles are alfo fimilar to one another.

For, because the Right Angle B DA is equal to the Right Angle ADC, and the Angle B A D has been proved equal to the Angle C; it follows, that the remaining Angle at B* fhall be equal to the remaining Cor. 31. 1.

Angle

Angle D A C. And fo the Triangle A B D is equi† 4 of this. angular to the Triangle A D C. Wherefore as + BD, fubtending the Angle B A D of the Triangle ABD, is to D A, fubtending the Angle at C of the Triangle ADC, which is equal to the Angle B AD; fo is AD, fubtending the Angle B of the Triangle ABD, to DC, fubtending the Angle D A C, equal to the Angle B. And, moreover, fo is BA to AC, fubtending the Right Angles at D; and, confequently, the Triangle ABD is fimilar to the Triangle A D C. Wherefore, if a Perpendicular be drawn in a Rightangled Triangle, from the Right Angle to the Bafe, then the Triangles on each Side of the Perpendicular are fimilar, both to the IVhole, and alfo to any another; which was to be demonftrated.

3. I. † 31. 1.

Coroll. From hence it is manifeft, that the Perpendi-
cular drawn in a Right-angled Triangle from the
Right Angle to the Bafe, is a mean Proportional
between the Segments of the Base. Moreover, ei-
ther of the Sides containing a Right Angle, is a
mean Proportional between the whole Bafe, and
the Segment thereof, which is next to the Side.

PROPOSITION IX.
PROBLEM.

To cut off any Part required from a given Right
Line.

LETAB be a Right Line given, from which muft
be cut off any required Part; fuppofe a third.

Draw any Right Line A C 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 equal to AD; join B C, and draw + D F through D, parallel to BC.

Then, because F D is drawn Parallel to the Side 12 of this. BC of the Triangle ABC, it fhall be, as CD is to DA, fo is BF to FA. But CD is double to D A. Therefore B F fhall be double to F A; and fo B A is triple to A F. Wherefore, there is cut off A F, a third Part required of the given Right Line AB; which was to be done.

PRO

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PROPOSITION X.

PROBLEM.

To divide a given undivided Right Line, as another given Right Line is divided.

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ETAB be a given undivided Right Line, and AC a divided Line. It is required to divide AB, as A C is divided.

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

31.1.

Then FH, H B, are each of them Parallelograms; and fo D H is + equal to F G, and H K to G B. And +34. 1. because H E is drawn parallel to the Side KC of the Triangle DK C, it fhall be as CE is to E D, fo is t 2 of this. KH to HD. But KH is equal to B G, and H D to GF. Therefore, as CE is to ED, fo is BG to GF. Again, becaufe FD is drawn parallel to the Side E G of the Triangle A G E, as ED is to D A, fo fhall + G F be to FA. But it has been proved, that C E is to ED, as B G is to GF. Therefore, as CE is to ED, fo is BG to GF; and as ED is to DA, fo is GF to FA. Wherefore, the given undivided Line A B is divided as the given Line AC is; which was to be done.

PROPOSITION XI.

PROBLEM.

Two Right Lines being given, to find a third
Proportional to them.

LE

ET 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 A B, AC, to the Points D and E; make BD equal to A C; join the Points B, C; and draw ‡ the Right Line D E thro' D, parallel to B. C. M

Then,

Then, because B C is drawn parallel to the Side 12 of this. DE of the Triangle ADE, it fhall be, as A'B is

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

Three Right Lines being given, to find a fourth
Proportional to them.

L

ET A, B, C, be three Right Lines given. It is required to find a fourth Proportional to them. Let DE and D F be two Right Lines, making any Angle E D F with each other. Now make D G equal to A, G Eequal to B, D H equal to C; and draw the Line GH, as alfo + E F thro' E, parallel to G H.

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Then, because G H is drawn parallel to E F, the
Side of the Triangle D E F, it fhall be, as DG is to
GE, fo is D H to HF. But D G is equal to A,
GE to B, and D H to C. Confequently, as A is to
B, fo is C to HF. Therefore, the Right Line HF, a
fourth Proportional to the three given Right Lines A, B,
C, is found; which was to be done.

PROPOSITION XIII.
PROBLEM.

To find a mean Proportional between two given
Right Lines.

LET the two given Right Lines be A B, BC. It is required to find a mean Proportional between them. Place A B, BC, in a direct Line; and on the Whole AC defcribe the Semicircle A DC, and I draw B D at Right Angles to AC from the Point B; and let A D, DC, be joined.

Then, becaufe the Angle ADC, in a Semicircle, isa Right Angle; and fince the Perpendicular DB is drawn from the Right Angle to the Bafe; therefore

DB

DB is a mean Proportional between the Segments of 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, baving one" Angle of the
one equal to one Angle of the other, have the
Sides about the equal Angles reciprocal; and
thofe 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 AB, BC, be equal Parallelograms, having

the Angles at B equal; and let the Sides D B, B E, be in one strait Line; then alfo I will the Sides 14. 1. FB, B G, be in one ftrait Line. I fay, the Sides of the Parallelograms AB, BC, that are about the equal Angles, are reciprocal; that is, as AD is to B E, fo is GB to B F.

For, let the Parallelogram F E be compleated.

I 1 of this.

Then, because the Parallelograms A B is equal to the Parallelogram B C, and FE is some other Parallelogram; it fhall be, as A B is to F E, fo is + BC † 7. 5. to FE; but as A B is to F E, fo is BD to BE; and as B C is to FE, fo is G B to B F. Therefore as D B is to BE, fo 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 BE, as GB ir to BF, I fay, the Parallelogram A B is equal to the Parallelogram B C.

For, fince D B is to B E, as GB is to B F; and DB to B E, as the Parallelogram A B to the Parallelogram F E; and G B to B F, as the Parallelogram B C to the Parallelogram F E; it fhall be as AE is to FE, fo is BC+ to F E. Therefore the Parallelogram A B is equal to the Parallelogram B C. And fo, equal Parallelograms, having one Angle of the

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