EXERCISES

ON PROPOSITIONS 18 AND 19.

1. The hypotenuse is the greatest side of a right-angled triangle.

2. If two angles of a triangle are equal to one another, the sides also, which subtend the equal angles, are equal to one another. Prop. 6. Prove this indirectly by using the result of Prop. 18.

3. BC, the base of an isosceles triangle ABC, is produced to any point D; shew that AD is greater than either of the equal sides.

4. If in a quadrilateral the greatest and least sides are opposite to one another, then each of the angles adjacent to the least side is greater than its opposite angle.

5. In a triangle ABC, if AC is not greater than AB, shew that any straight line drawn through the vertex A and terminated by the base BC, is less than AB.

6. ABC is a triangle, in which OB, OC bisect the angles ABC, ACB respectively: shew that, if AB is greater than AC, then OB is greater than OC.

ON PROPOSITION 20.

7. The difference of any two sides of a triangle is less than the third side.

8. In a quadrilateral, if two opposite sides which are not parallel are produced to meet one another; shew that the perimeter of the greater of the two triangles so formed is greater than the perimeter of the quadrilateral.

9. The sum of the distances of any point from the three angular points of a triangle is greater than half its perimeter.

10. The perimeter of a quadrilateral is greater than the sum of its diagonals.

11. Obtain a proof of Proposition 20 by bisecting an angle by a straight line which meets the opposite side.

ON PROPOSITION 21.

12. In Proposition 21 shew that the angle BDC is greater than the angle BAC by joining AD, and producing it towards the base.

13. The sum of the distances of any point within a triangle from its angular points is less than the perimeter of the triangle.

PROPOSITION 22. PROBLEM.

To describe a triangle having its sides equal to three given straight lines, any two of which are together greater than the third.

Let A, B, C be the three given straight lines, of which any two are together greater than the third.

It is required to describe a triangle of which the sides shall be equal to A, B, C.

Construction. Take a straight line DE terminated at the point D, but unlimited towards E.

Make DF equal to A, FG equal to B, and GH equal to C.

1. 3.

From centre F, with radius FD, describe the circle DLK. From centre G with radius GH, describe the circle MHK, cutting the former circle at K.

Join FK, GK.

Then shall the triangle KFG have its sides equal to the three straight lines A, B, C.

Proof. Because F is the centre of the circle DLK,

therefore FK is equal to FD:

but FD is equal to A ;

therefore also FK is equal to A.

Def. 11.

Constr.

Again, because G is the centre of the circle MHK,

therefore GK is equal to GH:

but GH is equal to C;

therefore also GK is equal to C.

And FG is equal to B.

Ax. 1.

Def. 11.

Constr.

Ax. 1.

Constr.

Therefore the triangle KFG has its sides KF, FG, GK equal

respectively to the three given lines A, B, C.

Q.E.F.

EXERCISE.

On a given base describe a triangle, whose remaining sides shall be equal to two given straight lines. Point out how the construction fails, if any one of the three given lines is greater than the sum of the other two.

PROPOSITION 23. PROBLEM.

At a given point in a given straight line, to make an angle equal to a given angle.

AA

Let AB be the given straight line, and A the given point in it; and let DCE be the given angle.

It is required to draw from A a straight line making with AB an angle equal to the given angle DCE.

Construction.

In CD, CE take any points D and E; and join DE.

From AB cut off AF equal to CD.

On AF describe the triangle FAG, having the remaining sides AG, GF equal respectively to CE, ED.

1. 3.

I. 22.

Then shall the angle FAG be equal to the angle DCE.

Proof.

Because

For in the triangles FAG, DCE,

FA is equal to DC,

and AG is equal to CE;

(and the base FG is equal to the base DE: Constr. therefore the angle FAG is equal to the angle DCE. I. S. That is, AG makes with AB, at the given point A, an angle

equal to the given angle DCE.

Q.E.F.

Constr.

Constr.

PROPOSITION 24.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one greater than the angle contained by the corresponding sides of the other; then the base of that· which has the greater angle shall be greater than the base of the other.

AA

B

G

F

Let ABC, DEF be two triangles, in which the two sides BA, AC are equal to the two sides ED, DF, each to each, but the angle BAC greater than the angle EDF :

then shall the base BC be greater than the base EF.

* Of the two sides DE, DF, let DE be that which is not greater than DF.

Construction. At the point D, in the straight line ED, and on the same side of it as DF, make the angle EDG equal to the angle BAC.

Make DG equal to DF or AC;
and join EG, GF.

I. 23.

I. 3.

Proof.

Then in the triangles BAC, EDG,

BA is equal to ED,

Hyp. Constr.

Because

and AC is equal to DG,

also the contained angle BAC is equal to the

contained angle EDG ;

Constr.

Therefore the triangle BAC is equal to the triangle EDG in all respects:

so that the base BC is equal to the base EG.

* See note on the next page.

I. 4.

Again, in the triangle FDG,
because DG is equal to DF,

therefore the angle DFG is equal to the angle DGF, 1. 5. but the angle DGF is greater than the angle EGF; therefore also the angle DFG is greater than the angle EGF; still more then is the angle EFG greater than the angle EGF. And in the triangle EFG,

because the angle EFG is greater than the angle EGF, therefore the side EG is greater than the side EF; 1. 19. but EG was shewn to be equal to BC; therefore BC is greater than EF.

Q.E.D.

* This condition was inserted by Simson to ensure that, in the complete construction, the point F should fall below EG. Without this condition it would be necessary to consider three cases: for F might fall above, or upon, or below EG; and each figure would require separate proof.

We are however scarcely at liberty to employ Simson's condition without proving that it fulfils the object for which it was introduced. This may be done as follows:

Let EG, DF, produced if necessary, intersect at K.
Then, since DE is not greater than DF,

that is, since DE is not greater than DG,

therefore the angle DGE is not greater than the angle DEG. But the exterior angle DKG is greater than the angle DEK: therefore the angle DKG is greater than the angle DGK. Hence DG is greater than DK.

But DG is equal to DF; therefore DF is greater than DK.

So that the point F must fall below EG.

I. 18. I. 16.

I. 19.