1.

EXERCISES.

ON THE IDENTICAL EQUALITY OF TRIANGLES.

Shew that the perpendiculars drawn from the extremities of the base of an isosceles triangle to the opposite sides are equal.

2. Any point on the bisector of an angle is equidistant from the arms of the angle.

3. Through O, the middle point of a straight line AB, any straight line is drawn, and perpendiculars AX and BY are dropped upon it from A and B shew that AX is equal to BY.

4. If the bisector of the vertical angle of a triangle is at right angles to the base, the triangle is isosceles.

5. If in a triangle the perpendicular from the vertex on the base bisects the base, then the triangle is isosceles.

6. If the bisector of the vertical angle of a triangle also bisects the base, the triangle is isosceles.

[Produce the bisector, and complete the construction after the manner of Theorem 8.]

7. The middle point of any straight line which meets two parallel straight lines, and is terminated by them, is equidistant from the parallels.

8. A straight line drawn between two parallels and terminated by them, is bisected; shew that any other straight line passing through the middle point and terminated by the parallels, is also bisected at that point.

9. If through a point equidistant from two parallel straight lines, two straight lines are drawn cutting the parallels, the portions of the latter thus intercepted are equal.

10. In a quadrilateral, ABCD, if ABAD, and BC=DC: shew that the diagonal AC bisects each of the angles which it joins; and that AC is perpendicular to BD.

11. A surveyor wishes to ascertain the breadth of a river which he cannot cross. Standing at a point A near the bank, he notes an object B immediately opposite on the other bank. He lays down a line AC of any length at right angles to AB, fixing a mark at the middle point of AC. From C he walks along a line perpendicular to AC until he reaches a point D from which O and B are seen in the same direction. He now measures CD: prove that the result gives him the width of the river.

ON THE IDENTICAL EQUALITY OF TRIANGLES.

Three cases of the congruence of triangles have been dealt with in Theorems 4, 7, 17, the results of which may be summarised as follows:

Two triangles are equal in all respects when the following three parts in each are severally equal:

1. Two sides, and the included angle.

Theorem 4.

Theorem 7

3. Two angles and one side, the side given in one triangle CORRESPONDING to that given in the other.

Theorem 17.

Two triangles are not, however, necessarily equal in all respects when any three parts of one are equal to the corresponding parts of the other.

For example:

(i) When the three angles of one are equal to the three angles of the other, each to each, the adjoining diagram shews that the triangles need not be equal in all respects.

(ii) When two sides and one angle in one are equal to two sides and one angle of the other, the given angles being opposite to equal sides, the diagram below shews that the triangles need not be equal in all respects.

=

F

For if AB = DE, and AC DF, and the ABC = the DEF, it will be seen that the shorter of the given sides in the triangle DEF may lie in either of the positions DF or DF'.

ACB,

NOTE. From these data it may be shewn that the angles opposite to the equal sides AB, DE are either equal (as for instance the DF'E) or supplementary (as the LACB, DFE); and that in the former case the triangles are equal in all respects. This is called the ambiguous case in the congruence of triangles. [See Problem 9, p. 82.]

If the given angles at B and E are right-angles, the ambiguity dis appears. This exception is proved in the following Theorem.

THEOREM 18.

Two right-angled triangles which have their hypotenuses equal, and one side of one equal to one side of the other, are equal in all respects.

Let ABC, DEF be two right-angled triangles, in which
the LABC, DEF are right angles,

the hypotenuse AC the hypotenuse DF,
and AB = DE.

It is required to prove that the ABC, DEF are equal in all

respects.

Proof.

Apply the AABC to the ADEF, so that AB falls on the equal line DE, and C on the side of DE opposite to F. Let C' be the point on which C falls.

Then DEC' represents the ABC in its new position.

Since each of the

DEF, DEC' is a right angle,

.. EF and EC' are in one straight line.

And in the AC'DF, because DF = DC' (i.e. AC),

(the

.. the DFC' the DC'F.

=

Hence in the A' DEF, DEC',

Theor. 5.

because the DFE the DC'E,

=

and the side DE is common

Proved.

the ADEF, DEC' are equal in all respects; Theor. 17.

that is, the A' DEF, ABC are equal in all respects.

Q.E.D.

*THEOREM 19. [Euclid I. 24.]

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

AA

CE

/K G

Let ABC, DEF be two triangles, in which

BA= ED,

and AC = DF,

but the BAC is greater than the EDF.

It is required to prove that the base BC is greater than the

base EF.

Proof. Apply the ABC to the DEF, so that A falls on D, and AB along DE.

Then because AB = DE, B must coincide with E.
Let DG, GE represent AC, CB in their new position.
Then if EG passes through F, EG is greater than EF;
that is, BC is greater than EF.

But if EG does not pass through F, suppose that DK bisects the FDG, and meets EG in K. Join FK.

because

Then in the ▲ FDK, GDK,

FD = GD,

DK is common to both,

and the included ▲ FDK = the included ▲ GDK;

Now the two sides EK, KF are greater than EF;
that is, EK, KG are greater than EF.
... EG (or BC) is greater than EF.

Q.E.D.

Conversely, if two triangles have two sides of the one equal to two sides of the other, each to each, but the base of one greater than the base of the other; then the angle contained by the sides of that which has the greater base, is greater than the angle contained by the corresponding sides of the other.

but the base BC is greater than the base EF.

It is required to prove that the BAC is greater than the LEDF.

Proof. If the BAC is not greater than the EDF,

L

it must be either equal to, or less than the EDF.

Now if the BAC were equal to the

EDF,

then the base BC would be equal to the base EF;

but, by hypothesis, it is not.

Theor. 4.

Again, if the ▲ BAC were less than the EDF, then the base BC would be less than the base EF; Theor. 19. but, by hypothesis, it is not.

That is, the BAC is neither equal to, nor less than the EDF; .. the BAC is greater than the EDF.

Q.E.D.

* Theorems marked with an asterisk may be omitted or postponed at the discretion of the teacher.