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Proposition 42. Theorem.

168. The perpendicular bisectors of the three sides of a triangle are concurrent.

Hyp. Let ABC be a ▲ ; D, E, F, the middle pts. of its sides; and DO, EO, FH, the 1s erected at D, E, F.

To prove that these Is meet in

a pt.

Proof. The two Is OD, OE,

since they cannot be ||, will meet at the pt. O.

Join OA, OB, OC.

bisector OD,

=

bisector OE,

E

'H

Because O is in the

... OA OB.

(66)

Because O is in the

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that is, O must lie on the bisector HF.

.. the three bisectors meet at the pt. O. Q.E.D. 169. COR. The point of intersection of the perpendicular bisectors, is equally distant from the three vertices of the triangle.

EXERCISES.

1. If the triangle in Prop. 41 is equilateral, find the value of the angle AOB.

2. If the same triangle is isosceles, and the angle Cis three times as great as either of the angles A and B, find the value of the angle AOB.

Proposition 43. Theorem.

170. The perpendiculars from the vertices of a triangle to the opposite sides are concurrent.

Hyp. Let ABC be a A, and R
AD, BE, CF the three Is from
A, B, C to the opp. sides.

To prove that these Is meet in a pt.

Proof. Through A, B, C, draw RQ, RP, PQ || respectively

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Similarly, RA = AQ, and PC = CQ;

that is, A, B, C, are the mid. pts. of QR, RP, PQ. Since AC is to PR, and BE is to AC,

... BE is also to PR.

A line to one of two || s is 1 to the other (71).

Similarly, AD and CF are

being the

to RQ and PQ. .. these three Is meet in a pt.,

bisectors of the three sides of the ▲ PQR (168). Q.E.D.

171. DEF. The intersection of the perpendiculars from the vertices of a triangle to the opposite sides is called its orthocentre.

The triangle formed by joining the feet of the perpendiculars is called the pedal or orthocentric triangle,

Proposition 44.. Theorem.

172. The three medial lines of a triangle are concurrent in a point of trisection,* the greater segment in each being towards the angular point.

Hyp. Let ABC be a ▲ ; D, E, F,

the mid. pts. of BC, AC, AB; and AD, BE, CF, the three medial lines.

To prove that these lines meet in the pt. that trisects them.

Proof. Let the two medials BE

and CF meet in O. Bisect BO in H, and CO in K.

[graphic]

Join HK, KE, EF, FH.

In the

and CO,

BOC, because H and K are the mid. pts. of BO

.. HK is || and to BC.

(155)

Also, in the ▲ ABC, because E and F are the mid. pts.

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That is, the medial BE cuts the medial CF at a pt. O,

one-third the way from F to C.

In the same way it may be proved that the medial AD cuts the medial CF at a pt. one-third the way from F to C; that is, at the same pt. O.

... The three medials are concurrent in the point that trisects them.

Q.E.D.

NOTE.-The point of intersection of the three medians of a triangle is called the centroid.

*When a line is divided into three equal parts it is said to be trisected,

SYMMETRY.

Symmetry with respect to an axis.

173. Two points are said to be symmetrical with respect to a straight line, when the straight line bisects at right angles the straight line joining the two

points.

Thus, the two points P and P' are
symmetrical with respect to the line
MN, if MN bisects PP' at right angles. M
The straight line MN is called the

axis of symmetry.

If we take the plane containing the

N

pt. P, and turn it about the axis MN, until the upper part is brought down on the part below MN, the line AP will take the direction AP', and the point P will coincide with the point P'. Thus, when two points are symmetrical with respect to an axis, if one of the parts of the plane be revolved about the axis to bring it down on the other part, the symmetrical points coincide.

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174. Two figures are said to be symmetrical with respect to an axis, when every point in one figure has its symmetrical point in the other.

Thus, the figures ABC, A'B'C' are symmetrical with respect to the axis MN, if every point in the figure ABC has a symmetrical point in A'B'C' with respect to MN.

In all cases, two figures that are symmetrical with re

spect to an axis, can be applied one to the other, by revolving either about the axis; consequently they are equal.

The corresponding symmetrical lines of symmetrical figures are called homologous lines. Thus, in the symmetrical figures, ABC, A'B'C', the homologous lines are AB and A'B', BC and B'C', AC and A'C'.

Symmetry with respect to a point.

175. Two points are said to be symmetrical with respect to a third point, when this third point bisects the straight line joining the two points.

Thus, P and P' are symmetrical with

respect to A, if the straight line PP' is P bisected at A.

The point A is called the centre of symmetry.

A

176. Two figures are said to be symmetrical with respect to a centre, when every point in one figure has its symmetrical point in the other.

Thus, the figures ABC, A'B'C' are symmetrical with respect to the centre O, if every point in the figure ABC has a symmetrical point in A'B'C'.

C

A

177. A figure is symmetrical with respect to an axis, when it can be divided by that axis into two figures symmetrical with respect M

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B

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