to BH, and also AF to FB; therefore FA and AG are equal to FB and BH, each to each; and the angle FAG has been proved equal to the angle FBH; therefore the base GF is equal to the base FH (I. 4): but it was proved that GE is equal to EH, and EF is common; therefore GE, EF are equal to HE, EF, each to each; and the base GF is equal to the base FH; therefore the angle GEF is equal to the angle HEF (I. 8); and consequently each of these angles is a right angle (I. Def. 10). Therefore FE makes right angles with GH, that is, with any straight line drawn through E in the plane passing through AB, CD. In like manner, it may be proved, that FE makes right angles with every straight line which meets it in that plane. But a straight line is at right angles to a plane when it makes right angles with every straight line which meets it in that plane (Def. 3): therefore EF is at right angles to the plane in which are AB, CD: which was to be proved.

PROP. V. PROB.

From a given point without a given plane to draw a straight line perpendicular to the plane.

Let A (fig. 12) be the given point without the plane MN: it is required to draw from the point A a straight line perpendicular to the plane MN.

DE

In the plane MN take any point B, and join AB; then if AB is perpendicular to the plane, what is required is now done. But if not, a straight line BC may be drawn in the plane MN not at right angles to AB (Def. 3). From A, in the plane ABC, draw AD perpendicular to BC (I. 12); and from D, draw, in the plane MN, perpendicular to BD (I. 11). Then if AD be at right angles to DE, it is at right angles to the plane EDB or MN (Prop. 4), and is the perpendicular required. But if not, from A, draw, in the plane ADE, AF perpendicular to DE (I. 12): AF is the perpendicular required.

Join FB. And because the angle ADB is a right angle, the square of AB is equal to the squares of AD and DB (I. 47); and because the angle AFD is a right angle, the square of AD is equal to the squares of AF and FD (I. 47): therefore the square of AB is equal to the squares of AF, FD and DB. of AF, FD and DB. But the square of FB is equal to the squares of FD and DB, because the angle FDB is a right angle: consequently the square of AB is equal to the squares of AF and FB; and therefore the angle AFB is a right angle (I. 48). And because AF is at right angles to DF and BF at their point of intersection, it is at right angles to the plane BFD (Prop. 4), that is, to the plane MN; and it is drawn from the given point A which was required to be done.

PROP. VI. THEOR.

From the same point without a plane, there can be only one perpendicular to the plane.

For, if possible, let the two straight lines AB and AC (fig. 13), meeting the plane MN in the point B and C, be both perpendicular to the plane; and join BC.

Because AB is perpendicular to the plane MN, and CB in that plane meets it, the angle ABC is a right angle (Def. 3); and because AC is perpendicular to the plane MN, and BC in that plane meets it, the angle ACB is a right angle. Consequently the two angles ABC, ACB of the triangle BAC are two right angles, which is impossible (I. 17). Therefore from the same point, &c.: which was to be proved.

PROP. VII. PROB.

From a given point in a given plane to draw a straight line perpendicular to the plane.

Let A (fig. 14) be the given point in the given plane MN: it is required to draw from the point A a straight line perpendicular to the plane MN.

From any point B without the plane MN draw BC perpendicular to it (Prop. 5) and meeting it in C; and join AC.

At A in the plane BCA draw AD at right angles to CA (I. 11): AD is perpendicular to the plane MN.

At C in the plane MN draw CE at right angles to AC; and join AE. Draw CD, in the plane BCA, to meet AD: and join DE.

Because BC is perpendicular to the plane MN, the angle BCE is a right angle (Def. 3): and because EC is at right angles to CB and CA, it is at right angles to the plane BCA (Prop. 4), and therefore to the straight line CD which meets it in that plane (Def. 3). And because the angle DCE is a right angle, the square of DE is equal to the squares of DC and CE (I. 47): but the square of DC is equal to the squares of DA and AC, because the angle DAC is a right angle; therefore the square of DE is equal to the squares of DA, AC and CE: and the square of AE is equal to the squares of AC and CE, because ACE is a right angle: consequently the square of DE is equal to the squares of DA and AE; and therefore the angle DAE is a right angle (I. 48). And because DA is at right angles to the two straight lines CA, EA at their point of intersection A, it is at right angles to the plane MN in which they are (Prop. 4); and it is drawn from the given point A; which was required to be done.

PROP. VIII. THEOR.

From the same point in a plane, there can be only one perpendicular to the plane, upon the same side of it.

For if it be possible, let the two straight lines AB, AC (fig. 15) be perpendicular to the plane MN, from the same point A, and upon the same side of it; and let the common section of the plane passing through BA and CA be the straight line DAE (Prop. 3); so that BĂ, CA and DAE are in one plane. And because BA is perpendicular to the plane MN, and EA meets it in that plane, the angle BAE is a right angle (Def. 3). For the same reason the angle CÃE is a right angle. Wherefore the angle BAE is equal to the angle CAE, in the same plane; the greater to the less, which is impossible. Therefore from the same point, &c.: which was to be proved.

PROP. IX. THEOR.

The perpendicular let fall from a given point to a given plane is the shortest straight line that can be drawn from the point to the plane.

Let A (fig. 16) be the given point, MN the given plane, and AB the perpendicular from A to the plane MN: AB is the shortest line that can be drawn from A to the plane MN.

For from A to the plane MN draw any other line AC, and join CB. Then since AB is perpendicular to the plane MN, the angle ABC is a right angle (Def. 3); and therefore the angle ACB is less than a right angle (I. 17). Consequently AB is less than AC (I. 19): which was to be proved.

Scholium. The perpendicular AB measures the distance of the point A from the plane MN.

PROP. X. PROB.

Through a given point to draw a plane perpendicular to a given

straight line.

Let AB be the given straight line, and O the given point (figs. 17,18). First, let the point O be in the straight line AB (fig. 17).

In two planes AOC, AOD passing through AB, draw OC, OD at right angles to AB (I. 11). Then AO being at right angles to each of the straight lines OC, OD at their point of intersection O, it is perpendicular to the plane MON passing through these lines (Prop. 4). Consequently the plane MON is perpendicular to the straight line AB (Def. 3); and it is drawn through the given point 0: which was required to be done.

Secondly, let the point O be without the straight line AB (fig. 18). From the point O, draw OC at right angles to AB (I. 12); and from the point C, in a plane passing through AB, draw CD also at

right angles to AB (I. 11). Then AB being at right angles to each of the lines CO, CD at their point of meeting C, it is perpendicular to the plane MON passing through these lines (Prop. 4). Consequently the plane MON is perpendicular to the straight line AB (Def. 3); and it is drawn through the given point 0: which was required to be done.

PROP. XI. THEOR.

Through a given point, only one plane can be drawn perpendicular to a given straight line.

First, let the given point O be in the given straight line AB (fig. 17): then besides the plane MON, drawn as in the last proposition, no other plane can be drawn through O, perpendicular to AB.

For, if possible, let another plane as MOR be perpendicular to AB, then a plane passing through AB may be drawn, intersecting the planes MN, MR in two straight lines as OE and OF; and consequently each of the angles AÕE, AOF, in the same plane, is a right angle (Def. 3); and they are equal to each other (Î. Ax. 11), which is impossible (I. Ax. 9).

Secondly, let the point O be without the given straight line AB (fig. 18) then besides the plane MON, drawn as in the last proposition, no other plane can be drawn through O perpendicular to AB.

For if another plane as MOR were perpendicular to AB, then a plane passing through AB may be drawn intersecting the planes MN, MR in two straight lines as OC and OE; and consequently each of the angles ACO, AEO would be a right angle (Def. 3), which is impossible (I. 16).

Therefore through a given point, &c.: which was to be proved.

PROP. XII. THEOR.

If three straight lines meet all in one point, and a straight line stand at right angles to each of them in that point; these three straight lines are in one and the same plane.

Let the straight line AB (fig. 19) stand at right angles to each of the straight lines BC, BD, BE, in B the point where they meet: BC, BD, BE are in one and the same plane.

If not, let, if it be possible, BD and BE be in one plane, and BC be above it; and let a plane pass through AB, BC, the common section of which, with the plane in which BD and BE are, is a straight (Prop. 3) line; let this be BF: therefore the three straight lines, AB, BC, BF, are all in one plane, viz. that which passes through AB, BC and because AB stands at right angles to each of the straight lines BD, BE, it is also at right angles (Prop. 4) to the plane passing through them; and therefore makes right angles with every straight line, in that plane, which meets it (Def. 3): but

BF, which is in that plane, meets it; therefore the angle ABF is a right angle but the angle ABC, by the hypothesis, is also a right angle; therefore the angle ABF is equal to the angle ABC, and they are both in the same plane, which is (I. Ax. 9) impossible; therefore the straight line BC is not above the plane in which are BD and BE: wherefore the three straight lines BC, BD, BE are in one and the same plane which was to be proved.

Cor. From this it follows, that all straight lines which are perpendicular to a given straight line, at a given point, are in the plane which is perpendicular to the given line, at that point.

Let the straight lines BC, BD, BE, BG, &c. be straight lines perpendicular to the straight line AB, at the point B: BC, BD, BE, BG, &c. are in the plane which is perpendicular to AB, at the point B.

For the line AB being at right angles to each of the lines BC, BD, BE, at their point of meeting, BC, BD, BE, are by this proposition in the same plane. Similarly BD, BE, BG are in the same plane. Consequently all the lines BC, BD, BE, BG are in the same plane. And since AB is perpendicular to the plane in which are BC, BD, BE, BG, &c. (Prop. 4), this plane is perpendicular to AB (Def. 3).

PROP. XIII. THEOR.

If a plane be perpendicular to a straight line at its middle point, every point in the plane will be equally distant from the extremities of the line.

Let the plane MN (fig. 20) be perpendicular to the straight line AB, at its middle point C; any point H in the plane MN is equally distant from A and B.

For join HC, HA, HB. Then because BC is equal to CA, and CH common to the two triangles BCH, ACH; and the angles BCH, ACH are right angles (Def. 3), BH is equal AH: which was to be proved.

Cor. Any point I, without the plane MN, is unequally distant from B and A. For BI is less than BH and HI (I. 20), that is, than AH and HI, or than AI.

PROP. XIV. THEOR.

Of straight lines drawn from a point without a plane, oblique to the plane, those are equal which are equally distant from the perpendicular drawn from the same point to the plane; and that which is nearer to the perpendicular is less than the one more remote. Also, conversely, equal straight lines drawn from a point without a plane, oblique to the plane, are equally distant from the perpendicular drawn from the same point to the plane; and of unequal straight lines drawn from the same point to the plane, the less is nearer to the perpendicular than the greater,