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.

C

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

Then from A a straight line can be drawn 1 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.

(I. 5)

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

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

.. LDGA is = ¿ EGA ;

.. AG is to BC.

(I. I)

PROPOSITION XII.

There cannot be drawn more than one perpendicular to a given straight line from a given point without it.

Let A be the given point, and BC the given straight line; and let AQ be drawn from A to BC. (I. 10) Then no other straight line besides AQ can be drawn from A to BC.

Draw any other straight line AR from A to BC. Produce AQ, and make the produced part QF= AQ. Join FR.

Then. AQR is a right 4,.. 4 FQR is a right . (1.9) Hence in the AS AQR, FQR,

AQ, QR and 4 AQR are respectively equal to FQ, QR, and FQR;

.. LARQL FRQ.

=

(I. I)

.. if ARQ were a right angle, the S ARQ, FRQ would be together two right angles, LS and .'. ARF would be a straight line; (I. 10) and thus two straight lines would inclose a space, which is impossible.

.. AR is not 1 to BC.

Similarly it may be proved that no other straight line besides AQ, drawn from A to BC, is 1 to it.

INEQUALITIES.

PROPOSITION XIII.

If a side of a triangle be produced the exterior angle is greater than either of the interior and opposite angles.

Let the side BC of a ABC be produced to D. Then shall the ▲ ACD be > either of the 4s BAC, ABC.

Bisect AC in E; join BE and produce BE to F, making EF-BE.

Join CF.

Then, AE, EB are respectively CE, EF,

and ▲ AEB is = ▲ CEF;

(1. 6)

Similarly it may be shewn, if AC be produced to G;

Hence it follows that

Any two angles of a triangle are together less than two right angles.

B

Let ABC be a A.

Then shall any two of its angles, as ABC and ACB, be together two right angles.

<

Produce BC to D.

Then ABC is the exterior

ACD; (1. 13)

.. Ls ABC, ACB are together < the 4s ACD, ACB, and are ... < than two right angles.

(1.9)

PROPOSITION XIV.

The greater side of every triangle has the greater angle opposite to it.

B

Let ABC be a ▲ having the side AC greater than the side AB.

Then shall the ABC be greater than the ACB.

From AC cut off AD = AB,
and join BD.

Then.. AB is = AD; .. ▲ ABD is = 1 ADB. (1. 2) And... the side CD of ▲ BCD has been produced to A, .. LADB is greater than ▲ DCB;

(I. 13)

.. also ▲ ABD is greater than ▲ DCB. Much more then is ABC greater than ▲ DCB,

i.e. LACB.