PROPOSITION XII. PROBLEM.

To draw a straight line perpendicular to a given straight line of unlimited length, from a given point without it.

Let AB be the given st. line, which may be produced any length both ways, and let C be the given pt. without it.

It is required to draw a st. line from CAB.

Д

F

H

E

B

Constr. Take any pt. D on the other side of AB, and from centre C at distance CD, describe O EGF meeting AB, produced if necessary, in F and G; (post. 3) bisect FG in H, (i. 10) and join CH, CF, CG.

Then the st. line CH shall be the given st. line AB. Dem. FH = HG (constr.), and HC is common to As FHC, GHC; the two sides FH, HC each to each; and base CF = base CG;

GH, HC,

(def. 15)

.. / FHC = ≤ GHC (i. 8), and these are adj. s. But, when a st. line standing on another st. line makes the adj. Zs equal to each other, each of these s is called a rt. ; and the st. line which stands on the other is called a to it.

(def. 10) Therefore from the given pt. C a perpendicular CH has been drawn to the given straight line AB. Q. E. F.

[If the straight line were not 'of unlimited length,' it might not meet the circle.]

If the perpendicular from the vertex of a triangle on the base bisect the base, the triangle is isosceles.

PROPOSITION XIII. THEOREM.

The angles which one straight line makes with another on one side of it, are either two right angles, or are together equal to two right angles.

Let the st. line AB make with CD, on one side of it, thes CBA, ABD.

Then these are either two rt. Zs, or are together two rt. Ls.

D

B

B

For, if CBA = ≤ DBA, each of them is a rt. .

(def. 10)

Constr. But if CBA be not = DBA, from B draw

BEL CD.

Dem. Then, Ls CBE, EBD are two rt. s.

(i. II) (def. 10)

And CBE = 2s CBA, ABE, add ▲ EBD to each; .. ▲s CBE, EBD = the three s CBA, ABE, EBD.

(ax. 2)

Again DBA = the two Ls DBE, EBA, add

▲ ABC to each;

.. Ls DBA, ABC: the threes DBE, EBA, ABC.

(ax. 2) But s CBE, EBD have been proved = the same three s.

.. 4s CBE, EBD = <s DBA, ABC; (ax. 1)

but CBE, EBD are two rt. ▲s; .. DBA, ABC are

[The learner should find out for himself the hypothesis and conclusion in this theorem, and in all which follow.]

DB meets the straight line ABC in B; BE, BF bisect the angles DBC, ABD. Show that FBE is a right angle.

PROPOSITION XIV. THEOREM.

If at a point in a straight line, two other straight lines, on the opposite sides of it, make the adjacent angles together equal to two right angles; these two straight lines shall be in one and the same straight line.

At the pt. B in the st. line AB, let the two st. lines BC, BD on the opposite sides of AB, make the adj. s ABC, ABD together = two rt. S.

Then BD shall be in the same st. line with CB.

A

B

For, if BD be not in the same st. line with CB, let BE be in the same st. line with it.

Dem. Then side of it,

AB makes with the st. line CBE, on one

s CBA, ABE; .. theses are together

two rt. Ls;

but s CBA, ABD are together

(i. 13)

two rt. s: (hyp.)

.. ▲ CBA, ABE = ▲s CBA, ABD;

(ax. II) take away from each of these equals ▲ CBA, and the ABE = rem. ▲ ABD:

(ax. 3)

rem. the less the greater, which is impossible; .. BE is not in the same st. line with CB.

And in the same way it may be demonstrated, that no other can be in the same st. line with it but BD, which therefore is in the same st. line with CB.

Therefore, if at a point, &c.

Q. E. D.

PROPOSITION XV. THEOREM.

If two straight lines cut one another, the vertical, or opposite angles shall be equal.

Let the two st. lines AB, CD cut one another at E. Then shall AEC = ▲ DEB, and Z CEB = L AED.

Dem. AE makes with CD at E s CEA, AED;

theses together

two rt. S.

(i. 13);

Again. DE makes with AB at E s AED, DEB; theses also together two rt. S.

but s CEA, AED have been shown together

rt. 2s;

(i. 13)

= two

../s CEA, AED = ▲s AED, DEB; take away from each of these equals the common ▲ AED, and remaining CEA = remaining DEB. (ax. 3) In the same way it may be demonstrated that CEB = AED.

Therefore, if two straight lines, &c.

Q. E. D.

Cor. 1. From this it is manifest, that if two st. lines cut each other, the Дs which they make at the pt. where they cut, together four rt. s.

=

Cor. 2. And consequently that all the ▲s made by any number of lines, meeting at one pt., together four

rt. s.

=

[The learner should work out the corollaries. No construction is required in this theorem: no more lines are needed than those specified in the particular enunciation.]