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Because the opposite sides of the figure P'C' are parallel,
and the angle C'a right angle, the figure is a rectangle ; and
PC being a section
of the prism SA'B'
by a plane parallel
to the base, it is
similar to the base, se
and is therefore a
rectangle. From si-
milar triangles SC':
SC = A'C: AC=P'B':PB; and SC': SC=B'C': BC=
P'A':PA; where P'B' and PB are the distances of the
point and its perspective from the horizontal plane, and
D'A', PA, their distances from the vertical plane.
COR.—Let the distances of the perspective of the point

from the horizontal and vertical planes be respectively
v and h, and those of the point itself o' and h', the dis-
tance of the picture d, and that of the point from the
primitive p, then
d+p:d=o':, and v=;

also, d +p:d=h' :h, and h=;

PROPOSITION III. To find the perspective of a point by means of a plane construction.

Let MR be the primitive, C its centre, CD the horizontal line, D the point of distance, and GR the ground line.

Produce the side KR, and make RA = P, the point's distance beyond the plane, RG its M_ distance from the vertical plane passing through KR, and RH its distance above the ground plane.

Make RB=RA; and join C, R, and D, B; through E draw EF pa- N G rallel to NR; join C, G; make PG perpendicular to NR and = RH; join C, P'; and draw FP parallel to GP'; and P is the perspective of the point.

For P' is evidently the seat of the point; and CP:PP' CF:FG=CE: ER=CD:BR=d:p. Hence (Pr. IV. 1) P is the perspective of the point. Cor. 1.-It is evident that the line EF is the locus of the

perspective of all the points in the line drawn on the ground plane parallel to the ground line, and at a dis

tance equal to RA from the primitive. Cor. 2.-The point F is evidently the perspective of the

seat of the given point on the ground plane. Schol.--The lines RN, RK, are sometimes called the scale of the front and the scale of heights, and CR is called the flying scale. Cor. 3.- The perspective of the point may be found

more simply thus :-Let CA bem
the distance of the point from the
horizontal plane, and AP its dis-
tance from the vertical plane, D
the point of distance, AP' paral-
lel to DC, and CA perpendicular
to it. Make P'F equal to the distance of the point
from the primitive; join P', C, and D, F, then P is
the perspective of P. For P' is evidently the seat of
the given point, and CP:PP' = CD:FP'; and there-
fore P is the perspective of the given point (Pr. IV.1).

PROPOSITION IV. The perspective of a plane figure parallel to the primitive is a similar figure, the dimensions of which are to the corresponding dimensions of the given figure in the ratio' of the distance of the picture to the sum of the distance lof the picture, and the distance of the point from the primitive.

Let S be the point of sight, and let the figure A'B'C' be parallel to MR, then it is evident (So. Ge. II. 14, Cor. 1)! that ABC is similar to A'B'C'; and that AC: A'C' = SC:' SC' =d:P + d. Cor. 1.-The projection of a straight line is a straight

line, unless it be directed to the eye, in which case it is na point." The projecting surface for any straight line as 'A'B' is' evidently a plane SA'B', the intersection AB of which with the primitive is a straight line.

When the line, as AD, is directed to S, its projection is the point A. Cor. 2.—The perspective of a

straight line parallel to the S
primitive is parallel to the

RA Cor. 3.-—The perspective of a

plane figure, whose plane passes through the point of

sight, is a straight line. For the projecting surfaces of its boundaries evidently lie in one plane.

PROPOSITION V. The projection of a circle inclined to the primitive is an ellipse, unless it be a subcontrary section of the projecting conical surface.

The perspective ACBD of the inclined circle A'C'B'D' is an ellipse, unless it be a subcontrary section.

For ACBD is a section of the pro M jecting conical surface SA'B', and it is therefore an ellipse, unless it be as subcontrary section, in which case it is a circle (Conic Sections IV.3).

PROPOSITION VI. The perspectives of parallel lines converge towards their vanishing point.

Let A'B' and C'D' be two parallel lines, and V their vanishing point, their pro- se jections AB and CD converge towards V.

ZB For since SV and A'B' are parallel, the projecting plane of A'B' will pass through SV, and therefore through V; and hence the perspective of A'B' passes through V. Similarly it is shown that the perspective of C'D' passes through V. Cor. 1.—The perspectives of lines perpendicular to the

primitive converge towards its centre. For the centre is their vanishing point. CoR. 2.—The perspectives of lines parallel to the primi

tive are parallel. For the vanishing point in this case is infinitely distant. It is evident also from Cor. 2 to Prop. IV. Cor. 3.—The perspectives of parallel horizontal lines con

verge towards a point in the horizontal line. Cor. 4.-The perspectives of parallel lines that are also

parallel to the vertical plane, converge towards a point 'in the vertical line. Cor 5.- The perspectives of horizontal lines, whose in

clination to the primitive is half a right angle, converge towards the point of distance.

PROPOSITION VI. Given the perspective of a straight line and its vanishing point, to find a point in it, which is the perspective of a point in the original line, that divides it in a given ratio.

Let AB be the given perspective, V the vanishing point, and P, Q, two lines in the given ratio.

From V draw any line VS in the plane of the primitive, and take any point S in it; m join S, A, and S, B, and produce these lines. Through A draw AH parallel to SV, and divide AH in G, so that AG: GH=P:Q; join S, G, and the point of intersection C is Z the point required.

For through C draw DE parallel to SV; then, from similar triangles AC: AV=DC: ŠV, and AC.SV = AV DC. Also ČB:BV = CE: SV, and CBSV = BV · CE. Also DC: CE =P:Q, and AV:BV=AV:BV; therefore (Pl. Ge. VI. 23, Cor. 1) AVDO: BY · CE=P• AV:Q. BV. But AC.SY:

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CB SV = AV DC:BVCE ; and therefore AC.SV: CBSV=PAV:Q.BV; or AC: CB=P: AV:Q.BV.

Let X and Y be the sides of two squares respectively equal to the given rectangles P• AV and Q.BV, and let Z be a third proportional to X and Y. Then AC: CB = X2: Y2 =X:Z. Hence divide AB in C in the ratio of X to Z.

The position of C, thus determined, is independent of the length and direction of SV; and therefore SV may be drawn of any length and in any plane. Hence let S be the point of sight on the farther side of the primitive MR, and SV will then be parallel to the original line, the extremities of which will lie in SA and SB produced. Let A'B' be the original line, then DE and AH are parallel to A'B' or SV; and these lines, with the given line AB, lie in one plane. Divide A'B' in C', so that A'C': C'B' =P:Q; then since DC : CE=P:Q, if SC and CC' be drawn, they will lie in one straight line, and therefore C is the perspective of C'.

PROPOSITION VII. Given the perspective of a straight line which is divided into segments having a given ratio and its vanishing point; to find those segments of its perspective, that are respectively the perspectives of the segments of the original line.

Let AB be the given perspective, and V the vanishing point, and X, Y, Z, three lines in the ratio of three segments into which the original line is divided.

Draw VS parallel to the ground line PR, and take in it any convenient point S; draw SA, SB, and produce them to E and F in PR. Divide M EF in G and H similarly to the original line; join S, G, and s, H; then AC, CD, DB, are the required segments.

OP EĠ HRT For (Pr. IV. 6), since SV and EF are parallel, and EF is similarly divided to the original line; therefore C, D, are the perspectives of the points of section, and consequently AC, CD, DH, are those of the segments of the original line.

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