PROPOSITION VII.

It is always possible to draw a straight line perpendicular to a given straight line from a given point in the same.

B D

Let BC be the given straight line,

and A the given point in it.

Then from A a straight line can be drawn 1 BC.

From AB, AC cut off equal parts AD, AE.

With D and E as centres, describe equal Os intersecting in F.

Join AF

· DA is = EA, DF=EF, and AF common to the two

PROPOSITION VIII.

From a given point in a given straight line there cannot be drawn more than one straight line perpendicular to it.

let AF and AG be each of them 1 to BC.

.. LS FAB, FAC are equal to one another, and also 4 s GAB, GAC are equal to one another, which is impossible.

.. from a given point in a given straight line there cannot be drawn more than one straight line to it.

COR. All right angles are equal to one another.

Let ABC, GHK be two right angles,

and let the ABC be applied to the GHK,
so that B may fall on H, and BC along HK,
then will BA fall along HG.

For if it fell in any other direction, as HX, then from H there would be drawn two straight lines HG, HX, 1 to HK, which is impossible;

and

C. G.

.. BA must fall along HG,

.. LABC is = L GHK.

PROPOSITION IX.

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

Let the straight line AB make with CD upon one side of it the LS ABC, ABD.

These shall either be two

then the ABC is the 4s CBE and EBA together; ... the S ABC and ABD are together the 4s CBE, EBA and ABD;

...

=

of which CBE is a right 4, and EBA, ABD make up a right

LS ABC, ABD are together

COR. If a number of straight lines BA, BC, BF, BE meeting in the point B form angles ABC, CBF, FBG, GBA which fill up the space about B, then these LS are together= 4 right L s.

For produce CB to X.

=

;

two rights.

PROPOSITION X.

If two straight lines, drawn from one extremity of a straight line on 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.

Let BC, BD drawn from B one extremity of AB make the adjacent s ABC, ABD together = two right angles.

Then shall BD be in the same straight line with BC.

For if not, if possible, let BE be in the same straight line with BC;

.. LS ABC, ABE are together = two right angles; (1.9) and are ..S ABC, ABD;

=

.. LABE is L ABD, which is impossible.

(hyp.)

... no other straight line than BD is in the same straight line with BC;

.. BD is in the same straight line with BC.

PROPOSITION XI.

It is always possible to draw a perpendicular to a given straight line of an unlimited length from a given point without it.

Let A be any point without the straight line BC of unlimited length.

Then from A a straight line can be drawn to BC.

With centre A describe a cutting BC in D and E. With centres D and E describe equal os intersecting. in F.

Join AF, cutting BC in G.
Then shall AG be 1 to BC.

Join AD, AE, FD, FE.

Then ·.· DA, AF, FD are respectively = EA, AF, FE; .. As DAF, EAF are equal in all respects,

Again,

=

and .. DAF is:

= LEAF.

(1. 5)

DA, AG and the included DAG are respectively EA, AG, and ▲ EAG;

... the as DAG, EAG are equal in all respects;

.. DGA is = LEGA;

.. AG is L to BC.

(I. I)