PROPOSITION XX. THEOREM.

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

Let ABC be A.

Then any two sides shall be together greater than the third side, namely BA, AC greater than BC; AB, BC greater than AC; and BC, CA greater than AB.

B

Constr. Produce any side BA to D, make AD = AC, (i. 3) and join DC.

Dem. AD = AC, (constr.).. 2 ADC = 2 ACD;

And. in A DBC, 4 BCD is greater than

BDC,

.. BD is greater than BC;

(i. 19)

but DB = BA and AC,

.. BA, AC are greater than BC.

In the same way it may be demonstrated that AB,

BC are greater than CA, also that BC, CA are greater than AB.

Therefore, any two sides, &c.

Q. E. D.

[Do not omit, before passing on, to work out the cases not demonstrated.]

1. Any three sides of a quadrilateral are together greater than the fourth.

2. The sum of the distances of any point from the angular points of a quadrilateral is greater than half the sum of the sides.

3. The sum of the diagonals of a quadrilateral is less than the sum of any four lines that can be drawn from any point whatever (except the intersection of the diagonals) to the four angles.

4. In rider 2 to prop. 15, show that the sum of the two straight lines is less than the sum of any other two lines drawn from the given points to any other point in the line.

5. How many triangles having two sides five feet and six feet long can be formed so that the third side shall contain a whole number of feet?

PROPOSITION XXI.

THEOREM.

If from the ends of a side of a triangle, there be drawn two straight lines to a point within the triangle; these shall be less than the other two sides of the triangle, but shall contain a greater angle.

Let ABC be A, and from the pts. B, C, the ends of the side BC, let the two st. lines BD, CD be drawn to a pt. D within the A.

Then BD and DC shall be less than BA and AC, the other two sides of the A, but shall contain

BDC greater

than

BAC.

E

D

Constr. Produce BD to meet AC at E.

Dem.

two sides of ▲ are together greater than the third side, (i. 20).. BA, AE are greater than BE;

to each of these unequals add EC;

.. BA, AC are greater than BE, EC. Again: CE, ED are greater than DC; to each of these unequals add DB;

(ax. 4)

(i. 20)

..CE, EB are greater than CD, DB. (ax. 4)

But it has been shown that BA, AC are greater than

BE, EC;

much more then are BA, AC, greater than BD, DC.

Again, ext. ▲ of A is greater than the int. opp.

(i. 16)

ext. BDC of A CDE is greater than ▲ CED; for the same reason ext. 4 CED of ▲ ABE is greater than < BAC;

and it has been demonstrated, that BDC is greater

than CEB;

much more then is BDC greater than

BAC.

Therefore, if from the ends of the side, &c. Q.E.D.

1. Three points A, B, C are taken inside a triangle PQR prove that the perimeter of the triangle ABC is less than that of the triangle PQR.

2. P is a given point, ABC a given triangle; show that twice the sum of PA, PB, PC is greater than the sum of the sides.

3. ABDE is a quadrilateral within the triangle ABC: the sides of the triangle are together greater than the sides of the quadrilateral.

PROPOSITION XXII.

PROBLEM.

To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third.

Let AP, BQ, CR be the three given st. lines, of which any two whatever are greater than the third, (i. 20) namely AP and BQ greater than CR; AP and CR greater than BQ; and BQ, CR greater than AP.

It is required to make ▲ of which the sides shall be equal to AP, BQ, CR, each to each.

A B C

R D

G

H

P

Constr. Take a st. line DE terminated at D, but unlimited towards E; make DF = AP, FG = BQ, and GH = CR;

(i. 3)

from centre F, at distance FD, describe the ODKL;

(post. 3)

and from centre G, at distance GH, describe O HLK; and join KF, KG.

Then shall ▲ KFG have its sides the three st. lines AP, BQ, CR.

Dem.

F is the centre of DKL, .. FD = FK;

but FD AP, (constr.) .. FK = AP.

(def. 15)