COROLLARY 1. Any two angles of a triangle are together less than two right angles.

For the LABC is less than the LACD:

to each add the LACB.

Proved.

Then the ABC, ACB are less than the LACD, ACB, therefore, less than two right angles.

4

B

COROLLARY 2. Every triangle must have at least two acute angles.

For if one angle is obtuse or a right angle, then by Cor. 1 each of the other angles must be less than a right angle.

COROLLARY 3. Only one perpendicular can be drawn to a straight line from a given point outside it.

If two perpendiculars could be drawn to AB from P, we should have a triangle PQR in which each of the LS PQR, PRQ would be a right angle, which is impossible.

1.

EXERCISES.

Prove Corollary 1 by joining the vertex A to any point in the

base BC.

2. ABC is a triangle and D any point within it. joined, the angle BDC is greater than the angle BAC.

(i) by producing BD to meet AC.

If BD and CD are
Prove this

(ii) by joining AD, and producing it towards the base.

3. If any side of a triangle is produced both ways, the exterior angles so formed are together greater than two right angles.

4. To a given straight line there cannot be drawn from a point out. side it more than two straight lines of the same given length.

5. If the equal sides of an isosceles triangle are produced, the exterior angles must be obtuse,

THEOREM 9. [Euclid I. 18.]

If one side of a triangle is greater than another, then the angle opposite to the greater side is greater than the angle opposite to the less.

Let ABC be a triangle, in which the side AC is greater than the side AB.

It is required to prove that the LABC is greater than the LACB. From AC cut off AD equal to AB.

Proof.

Join BD.

Because AB = AD,

.. the ABD = the ADB

Theor. 5.

But the exterior ADB of the ▲ BDC is greater than the interior opposite ▲ DCB, that is, greater than the ACB.

.. the ABD is greater than the ACB.

Still more then is the ABC greater than the ACB.

Q.E.D.

Obs. The mode of demonstration used in the following Theorem is known as the Proof by Exhaustion. It is applicable to cases in which one of certain suppositions must necessarily be true; and it consists in shewing that each of these suppositions is false with one exception: hence the truth of the remaining supposition is inferred.

THEOREM 10. [Euclid I. 19.]

If one angle of a triangle is greater than another, then the side opposite to the greater angle is greater than the side opposite to the less.

Let ABC be a triangle, in which the ABC is greater than the ACB.

It is required to prove that the side AC is greater than the side AB.

Proof.

If AC is not greater than AB,

it must be either equal to, or less than AB.

Now if AC were equal to AB,

then the ABC would be equal to the ACB; Theor.5.

but, by hypothesis, it is not.

Again, if AC were less than AB,

then the ABC would be less than the ACB; Theor. 9.

but, by hypothesis, it is not.

That is, AC is neither equal to, nor less than AB.

... AC is greater than AB.

[For Exercises on Theorems 9 and 10 see page 34.]

Q.E.D.

THEOREM 11. [Euclid I. 20.]

Any two sides of a triangle are together greater than the third side.

Let ABC be a triangle.

It is required to prove that any two of its sides are together greater than the third side.

It is enough to shew that if BC is the greatest side, then BA, AC are together greater than BC.

Produce BA to D, making AD equal to AC.

But BD BA and AC together;

.. BA and AC are together greater than BC.

Q.E.D.

NOTE. This proof may serve as an exercise, but the truth of the Theorem is really self-evident. For to go from B to C along the straight line BC is clearly shorter than to go from B to A and then from A to C. In other words

The shortest distance between two points is the straight line which joins them.

THEOREM 12.

Of all straight lines drawn from a given point to a given straight line the perpendicular is the least.

Let OC be the perpendicular, and OP any oblique, drawn from the given point Ō to the given straight line AB.

It is required to prove that OC is less than OP.

Proof. In the OCP, since the OCP is a right angle, .. the ▲ OPC is less than a right angle;

that is,

the OPC is less than the OCP.
... OC is less than OP.

Theor. 8. Cor.

Theor. 10.

Q.E.D.

COROLLARY 1. Hence conversely, since there can be only one perpendicular and one shortest line from O to AB,

If OC is the shortest straight line from O to AB, then OC is perpendicular to AB.

COROLLARY 2. Two obliques OP, OQ, which cut AB at equal distances from C the foot of the perpendicular, are equal.

The As OCP, OCQ may be shewn to be congruent by Theorem 4; hence OP=OQ.

COROLLARY 3. Of two obliques OQ, OR, if OR cuts AB at the greater distance from the foot of the perpendicular, then OR is

greater than OQ.

The LOQC is acute,.. the LOQR is obtuse ;

H.8.G.

.. the OQR is greater than the LORQ;

.. OR is greater than OQ

с