MODEL DEMONSTRATIONS

The angles opposite the equal sides of an isosceles triangle are equal.

Proof Suppose AX is drawn bisecting

ZBAC and meeting BC at X.

In the ABAX and CAX

AX AX (Identical).

:.^ ABX = ▲ ACX. (Two ▲ are = if two sides and the included of one are = respectively to two sides and the included of the other.) Hence, B = C. (Homologous parts of equal figures are equal.) Q.E.D.

Two triangles are equal if the three sides of one are equal respectively to the three sides of the other.

sides (BC and ST) coincide, and A is opposite ST from R. Draw RA. RS AS (Hypothesis).

=

AASR is isosceles. (An isosceles ▲ is a ▲ two sides of which are equal.) :. ≤ SRA = 2 SAR..... (1) . . (The opp. the sides of an isos. ▲ are =.) Again, TR = AT (Hypothesis).

▲ TRA is isosceles. (Same reason as before.)

< TAR... (2).. (Same reason as for (1).) Adding equations (1) and (2).

=

4 SRT = 2 SAT. (If 's are added to 's the results are =.) Consequently, the ^ RST = ▲ AST. (Two ▲ are = if two sides and the included of one are = respectively to two sides and the included of the other.)

That is, ▲ RST = ▲ ABC. (Substitution; A ABC is the same as ▲ AST.)

Q.E.D.

The preceding form of demonstration will serve to illustrate an excellent scheme of writing the proofs. It will be observed that the statements are at the left of the page and their reasons at the right. This arrangement will be found of great value in the saving of time, both for the pupil who writes the proofs and for the teacher who reads them.

61. The converse of a theorem is the theorem obtained by interchanging the hypothesis and conclusion of the original theorem. Consult 44 and 45; 79, 80, and others.

Every theorem which has a simple hypothesis and a simple conclusion has a converse, but only a few of these converses are actually true theorems.

For example: Direct theorem: "Vertical angles are equal."

Converse theorem: "If angles are equal, they are vertical." This statement cannot be universally true.

The theorem of 120 is the converse of that of 55.

62. Auxiliary lines. Often it is impossible to give a simple demonstration without drawing a line (or lines) not described in the hypothesis. Such lines are usually dotted for no other reason than to aid the learner in distinguishing the lines mentioned in the hypothesis and conclusion from lines whose use is confined to the proof. Hence, lines mentioned in the hypothesis and conclusion should never be dotted. (The figure used in 57 should contain no dotted line.)

63. Superposition. It is worthy of note that demonstration by the method of superposition is never employed unless the hypothesis gives a pair of equal angles.

64. Homologous parts. Triangles are proven equal in order that their homologous sides, or homologous angles, may be proven equal. This is a very common method of proving lines equal and angles equal.

65. The distance from one point to another is the length of the straight line joining the two points.

66. THEOREM. If lines be drawn from any point in a perpendicular erected at the midpoint of a straight line to the ends of the line,

I. They will be equal.

II. They will make equal angles with the perpendicular.

III. They will make equal angles with the line.

Given AB to CD at its

:

midpoint, B; P any point in AB; PC and PD.

To Prove: I. PC=PD;

II. ≤ CPB = ≤ DPB ; and

III. Z C = L D.

Proof: In rt. A PBC and

PBD, BC = BD (Hyp.); BP = BP (Iden.).

. . ^ PBC = ▲ PBD. (Why ?) (53.)

B

I. PC = PD (Why?) (27;) II. ≤ cpB = ≤ DPB (Why ?); III. ≤C=≤D (Why?).

Q.E.D.

67. THEOREM. Any point in the perpendicular bisector of a line is equally distant from the extremities of the line. (See 66, I.)

68. THEOREM. Any point not in the perpendicular bisector of a line is not equally distant from the extremities of the line.

DO+OP > PD. (Why?) (Ax. 12.) But co = od (67). .. CO+OP > PD. (Substitution; Ax. 6.)

That is, PC > PD, or PC is not = PD.

Q.E.D.

69. THEOREM. If a point is equally distant from the extremities of a line, it is in the perpendicular bisector of the line. (See 67 and 68.)

70. THEOREM. Two points each equally distant from the extremities of a line determine the perpendicular bisector of the line. Each point is in the bisector (69); two points determine a line (5).

71. THEOREM. Only one perpendicular can be drawn to a line from an external point.

Given: PRL to AB from P; PD any other line from P to AB.

To Prove: PD cannot be to AB; that is, PR is the only 1 to AB from P.

Proof: Extend PR to S, making RS PR; draw DS.

In rt. A PDR and SDR, PR = RS (Const.).

DR = = DR (Iden.). ..A PDR=

▲ SDR. (Why?) (53.)

A

R

B

../ PDR =

SDR (27). That is, PDR = half of ≤ PDs.

Now PRS is a straight line (Const.).

.. PDS is not a straight line (39).

../ PDS is not a straight angle (18).

../ PDR, the half of ▲ PDS, is not a right angle (36).

.. PD is not (17). .. PR is the only 1.

Q.E.D.

Ex. 1. Through how many degrees does the minute hand of a clock move in 15 min.? in 20 min.? Through how many degrees does the hour hand move in one hour? in 45 minutes? in 10 minutes?

Ex. 2. How many degrees are there in the angle between the hands of a clock at 9 o'clock? at 10 o'clock? at 12:30? at 2:15? at 3:45?

Ex. 3. THEOREM. If two lines be drawn bisecting each other, and their ends be joined in order, the opposite pairs of triangles will be equal. [Use 51 and 52.]

72. THEOREM. Two right triangles are equal if the hypotenuse and an adjoining angle of one are equal respectively to the hypotenuse and an adjoining angle of the other.

Given: Rt. A LMN and RST; LN = RT ; and ▲ L = /R. To Prove: ▲ LMN = ▲ RST.

Proof: Superpose ▲ LMN upon ▲ RST so that L coincides with its equal, R, LM falling along RS. Then LN will fall on RT and point N will fall exactly on T (LN = RT by Hyp.). Now NM and TS will both be L to RS from T (Rt. A by Hyp.). .. NM will coincide with TS (71).

... ▲ LMN = ^ RST (28).

Q.E.D.

73. THEOREM. Two right triangles are equal if the hypotenuse and a leg of one are equal respectively to the hypotenuse and a leg of the other.

Given: Rt. ▲ IJK and LMR ; KI = RM; KJ = RL.

To Prove: AIJK = ALMR.

Proof Place AIJK in the position of AXLR so that the equal sides, KJ and RL, coincide and I is at X, opposite RL from M.