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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.

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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,

and ... ▲ DAF is = LEAF

(1.5)

Again, ·.· DA, AG and the included DAG are re

spectively = EA, AG, and ▲ EAG;

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

..4 DGA is:

=

LEGA;

.. AG is 1 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.

A

R

C

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 QFAQ. Join FR.

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

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

LARQ = L FRQ.

.. if ARQ were a right angle,

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=

(I. I)

the SARQ, FRQ would be together two right angles, 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 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.

D

Let the side BC of ▲ ABC be produced to D.
ACD be> either of the 4S BAC, ABC.

Then shall the

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

Join CF.

Then, ·. AE, EB are respectively = CE, EF,

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But the ACD is > the ▲ ECF;

.. ▲ ACD is > the EAB; i.e. ▲ CAB.

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

that

BCG is the ABC.

But ACD is = LBCG;

(1.6)

.. LACD is the ABC.

Hence it follows that

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

D

Let ABC be a ▲.

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

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Produce BC to D.

Then ABC is < the exterior ACD; (1. 13) .. LS ABC, ACB are together the 4s ACD, ACB,

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and are .. < than two right angles.

(I. 9)

PROPOSITION XIV.

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

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

Then shall the ABC be greater than the ACB.

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And ⚫. the side CD of ▲ BCD has been produced to A,

(I. 13)

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

i.e. LACB.

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