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Since B A C is an isosceles triangle and A D bisects B C, A D is perpendicular to B C (El. (75) ).* Hence A D C and A D B are right angled triangles. Also E DC and E D B are right angles. Hence the square of EC is equal to the sum of the squares of ED and DC. But also because of the right angle EAD, the square of ED is equal to the sum of the squares of E A and A D. Hence the square of E C is equal to the sum of the squares of ED, AD, and DC. But because of the right angle ADC the square of A C is equal to the sum of the squares of A D and DC. Hence the square of E C is equal to the sum of the squares of E A and AC, and therefore the angle E A C is right.

In the same manner it may be proved that the angle E AB is right.

PROPOSITION VIII. (17) A right line which is perpendicular to two right

lines which intersect, is also perpendicular to every line in their plane drawn through their

point of intersection. For let A E be perpendicular to the lines A B and AC, and let A D be any other line through A in the plane of the angle BAC. Let any line BC be drawn intersecting the sides A B, AC, and bisected by the line AD, and draw EB, EC.

In the triangle B É C, since the line E D bisects the base B C, the sum of the squares of E B and EC is equal to twice the sum of the squares of ED and DC (El. (299)). Also because of the right angled triangles B AE and C AE, the square of B E is equal to the sum of the squares of B A and A E, and the square of CE is equal to the sum of the squares of C A and A E. Hence the sum of the squares of BE and CE is equal to the sum of the squares of BA and CA together with twice the square of E A. Hence twice the sum of the squares of D E and D C is equal to the sum of the squares of B A and AC together with twice the square of A E.

But since the line A D bisects B C, the sum of the squares of B A and C A is equal to twice the sụm of the squares of D A and DC. Hence twice the sum of the squares of D E and DC is equal to twice the sum of the squares of E A, AD, and D C. Take from both twice the square of D C, and we find that twice the sum of the squares of E A and AD is equal to twice the

* Where El. precedes a reference, the article referred to is in the first six books of Euclid's Elements; otherwise the reference is to the Solid Geometry.

square of E D, and therefore the sum of the squares of E A and DA is equal to the square of E D, and therefore the angle E AD is right; and the same may be proved of any other right line drawn through A in the plane of the angle B A C.

Hence the line A E is perpendicular to every right line drawn through A in the plane of the angle B A C. (18) Dec.-A right line, such as A E, which is drawn from a point in a plane so as to be perpendicular to all lines drawn in the plane through that point, is said to be perpendicular to the plane itself. (19) CoR.-Through the same point A in a given plane BAC only one right line A E can be drawn perpendicular to the plane. For if another could be drawn let it be A F, and let A C be the intersection of the plane of the angle FAE with the given plane BAC. Since FA and E A are both perpendicular to the plane BAC, the angles E A C and FA C are both right, and therefore equal, a part to the whole, which is absurd. (20) DEF.—We shall call that point at which a perpendicular to a plane meets the plane, the foot of the perpendicular.

PROPOSITION IX.

P

B

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D

(21) From a given point out of a given plane a right

line may be drawn perpendicular to the plane,

and only one such line can be drawn. Let A B C be the given plane, and P the given point. Draw any line PB from P to the plane; if this be perpendicular to the plane, the proposition is true. If not, let any line B C be drawn from the point B, and in the given plane, and let the angle PB C be acute. From P inflect on B C a right line PC, equal in length to PB, so that the triangle BPC shall be isosceles. Bisect B C at D, and in the given plane draw D A perpendicular to B C, and from P draw P A perpendicular to AD. This line P A will be perpendicular to the given plane.

For draw AB, AC, and PD.

Since BPC is an isosceles triangle and PD bisects the base, it is perpendicular to the base, and the angles P DB, PDC are right. The angles A D C and ADB are right by construction. The square of P B is equal to the sum of the squares of PD and D B. But since PAD is right by construction, the square of PD is equal to the sum of the squares of PA and A D. Hence the square of P B is equal to the sum of the squares of PA, AD, and B D. But since A D B is a right angle, the sum

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of the squares of A D and B D is equal to the square of AB. Hence the square of PB is equal to the sum of the squares of PA and A B, and therefore the angle PAB is right, and since PAD is also right, the line PA is perpendicular to the plane ABC (17, 18).

It is evident that only one perpendicular can be drawn from the same point P, because if two were supposed to be drawn they would · both be perpendicular to the line joining the points where they would meet the plane, and thus two right angles would be in the same triangle.

PROPOSITION X.

(22) Of the several lines which may be drawn from a

given point P to a given plane ABC, 1°. The perpendicular P A is the shortest. 2o. Those which are equally inclined to the per

pendicular are equal, and vice versa. 3o. The greater the angle which a line makes

with the perpendicular, the greater the line is,

and vice versa. 4° Lines which meet the plane at equal dis

tances from the foot of the perpendicular are

equal, and vice versa. 5o. The more distant the point where a line

meets the plane is from the foot of the perpendicular, the greater is the line, and vice

versá. 1o. The perpendicular PA is the shortest line because it is the side of a right angled triangle PA B, of which any p other line P B is the hypotenuse.

2°. If the angles CPA and B P A be equal, since the angles PAB and PAC are right, and P A common to the triangles CPA and B PÅ, the sides B P and PC are equal.

If the sides PB and PC be equal, since the angles PAB and PAC are right and PA common, the angles A P B and APC must be equal.

3o. If the angle A PB be greater than the angle A PC, since the side P A is common and the angles at A right, the side A B must be greater than A C, and therefore P B greater than PC.

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If PB be greater than P C, since PA is common and the angles at A right, the side A B is greater than A C, and therefore the angle A PB greater than the angle A PC.

4°. If A B be equal to A C, since AP is common, and the angles at A right, the lines P B and PC must be equal.

If PB be equal to PC, since P A is common, and the angles at A right, the sides A B and AC are equal.

5°. If A B be greater than A C, since P A is common, and the angles at A right, the line P B must be greater than P C.

If P B be greater than P C, since P A is common, and the angles at A right, the line A B must be greater than Ać. (23) Cor.—Hence all equal oblique lines from a point P to a given plane terminate in the circumference of a circle described upon the plane of which the foot of the perpendicular is the centre; and also all lines which drawn from the same point to a plane meet the plane in the circumference of a circle, of which the foot of the perpendicular is the centre, are equal.

PROPOSITION XI.

P

(24) If P A be perpendicular to the plane A B C, and

BC be a right line in that plane, and from A
A D be drawn perpendicular to BC, then PD

will be perpendicular to BC. On each side of D take equal parts D B and D C, and draw PB, PC, AB, and A C.

Since D B is equal to D C, and the angles at D are right, A B is equal to AC. Also since A B and AC are equal, and the angles PAB and PAC are right, P B and P C are equal. In the isosceles triangle BPC, PD bisects the base and is therefore perpendicular to it. (25) Cor.—It is evident that B C is perpendicular to the plane PD A since it is perpendicular to D P and D A. (26) The lines P A and B C are an instance of two lines which, without being parallel, can never meet if produced indefinitely, since the same plane cannot be drawn through them.

B

A

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PROPOSITION XII.

(27) If a right line be perpendicular to a plane, every

right line which is parallel to it is perpendicular to the same plane.

2

A

B

Let P A be perpendicular to the plane A B C, and let D E be parallel to PA. Draw A D, and draw B C perpendicular to AD, and in the given plane and from any point P in the perpendicular A P draw PD.

The line B C is perpendicular to the plane PDA (25), and since DE is parallel to P A the same plane may be drawn through them, and this plane is evidently that of the angle E D A. Since B C is perpendicular to the plane PD A or E DA, E D C is a right angle. But because of the parallels, E D A is a right angle. Hence E D is perpendicular to DC and DA, and is therefore perpendicular to the plane ABC.

PROPOSITION XIII.

(28) Perpendiculars to the same plane are parallel.

For if two perpendiculars be not parallel, through the foot of one draw a parallel to the other. This will be perpendicular to the plane (27), and therefore two perpendiculars to the same plane would pass through the same point, which cannot be (19).

PROPOSITION XIV.

(29) Right lines which are parallel to the same right

line are parallel to each other.

For the plane which is perpendicular to the last will be also perpendicular to the others (27), and therefore the other lines must be parallel to each other (28).

This proposition applied to parallels in the same plane is the thirtieth proposition of the first book of the Elements. (30) DeF.—When a plane and a right line are so placed that each being indefinitely produced they will never meet, they are said to be parallel.

PROPOSITION XV.

(31) If two right lines be parallel, every plane drawn

through one of them is parallel to the other.

(The plane of the parallels themselves is here excepted.)

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