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 4. (1.9) Hence in the As AQR, FQR,

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

.. LARQ= L FRQ.

(I. 1)

.. if ARQ were a right angle, the S ARQ, 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 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.

B

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

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

H

Join CF.

Then, AE, EB are respectively CE, EF,

=

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

that

BCG is the ABC.

But ACD is = L BCG;

(1.6)

.. LACD is > the ABC.

Hence it follows that

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

B

D

Let ABC be 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 = LADB. (1. 2)
And.. the side CD of a BCD has been produced to A,

.. LADB is greater than 4 DCB;

... also

ABD is greater than ▲ DCB.

(I. 13)

Much more then is ABC greater than DCB,

i.e. LACB.

PROPOSITION XV.

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

B

Let ABC be a ▲ having the ABC > the ACB,

then shall the side AC be > the side AB.

For if not, AC must either be

=

AB or < AB.

AB, for, then would ABC be = LACB,

(1. 2)

And AC is not < AB, for then would ▲ ABC be < LACB, but it is not.

(I. 14)

.. AC is > AB.