THEOREM 3

The angles at the base of an isosceles triangle are equal to each other.

Hypothesis.-ABC is an isosceles ▲ having AB = AC. To prove that LB = L C.

Hypothetical Construction.-Draw the st. line AD to

36. The two As ADB, ADC, in the diagram of Theorem 3, are congruent, and if the isosceles A be folded along the bisector of the vertical as crease, the parts on one side of the bisector will exactly fit the corresponding parts on the other side.

Definition. When a figure can be folded along a line so that the part on one side exactly fits the part on the other side, the figure is said to be symmetrical with respect to that line.

The line along which the figure is folded is called an axis of symmetry of the figure.

Hence the bisector of the vertical

celes is an axis of symmetry of the A.

of an isos

It follows from the above definition of a symmetrical figure that

If a figure is symmetrical with respect to a st. line, for every point on one side of this axis of symmetry there is a corresponding point on the other side.

Show by folding, in the diagram of Theorem 3, that if BC, the side AB = the side AC.

37.-Exercises

1. An equilateral ▲ is equiangular.

2. ABC is an equilateral A, and points D, E, F, are taken in BC, CA, AB respectively, such that BD = CE = AF. Show that DEF is an equilateral ▲.

3. Show that the exterior Ls at the base of an isosceles A are equal to each other.

4. The opposite s of a rhombus are equal to each other.

5. ABC is an isosceles ▲ having

AB = AC, and the base BC pro

duced to D and E such that BD = CE. Prove that ADE is an isosceles A.

6. AC, AD are two st. lines on opposite sides of AB. Prove that if the bisectors of LS BAC, BAD are at rt. Zs, AC, AD must be in the same st. line.

7. If a figure be symmetrical with respect to a st. line, the st. line joining any two corresponding points cuts the axis at rt. Ls.

SECOND CASE OF THE CONGRUENCE OF TRIANGLES

THEOREM 4

If two triangles have the three sides of one respectively equal to the three sides of the other, the two triangles are congruent.

Hypothesis.-ABC, DEF are two As having AB = DE, DF and BC = EF.

AC =

To prove that ▲ ABC= A DEF.

Proof.-Let A DEF be applied to ▲ ABC so that the vertex E falls on the vertex B and EF falls along BC. Then EF = BC, the vertex F falls on C. Let D take the position D' on the side of BC remote from A.

Note. In the proof of this theorem three cases may occur :-AD' may cut BC as in Fig. 1, or not cut BC as in Fig. 2, or pass through one end of BC as in Fig. 3.

The proof given above is that of the first case. The pupil should work out the proofs of the other two cases.

38.-Exercises

1. If the opposite sides of a quadrilateral be equal, the opposites are equal.

2. A diagonal of a rhombus bisects each of the LS through which it passes, and consequently, the diagonal is an axis of symmetry in the rhombus.

3. If in a quadrilateral ABCD the sides AB, CD be equal and ABC 4 BCD, prove that CDA = / DAB.

=

4. Show that equal chords in a circle subtend equal s at the centre.

5. Prove that the diagonals of a rhombus bisect each other at rt. 2s.

Imp

THEOREM 5

If two isosceles triangles are on the same base, the straight line joining their vertices is an axis of symmetry of the figure; and the ends of the base are corresponding points.

44

B

E

B

Hypothesis.-ABC, DBC are two isosceles As on the same base BC.

To prove that AD is an axis of symmetry of the figure.

Proof-AD, or AD produced, cuts BC at E.