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11. If in a diagonal of a parallelogram any two points equidistant from its extremities are joined to the opposite angles, the figure thus formed will be also a parallelogram.

12. ABC is a given equilateral triangle, and in the sides BC, CA, AB the points X, Y, Z are taken respectively, so that BX, CY and AŹ are all equal. AX, BY, CZ are now drawn, intersecting in P, Q, R : shew that the triangle PQR is equilateral.

13. If in the sides AB, BC, CD, DA of a parallelogram ABCD four points P, Q, R, S are taken in order, one in each side, so that AP, BQ, CR, DS are all equal; shew that the figure PQRS is a parallelogram.

14. In the figure of 1. 1, if the circles intersect at F, and if CA and CB are produced to meet the circles in P and Q respectively ; shew that the points P, F, Q are in the same straight line; and shew also that the triangle CPQ is equilateral. [Problems marked (*) admit in general of more than one solution.]

15. Through two given points draw two straight lines forming with a straight line given in position, an equilateral triangle.

*16. From a given point it is required to draw to two parallel straight lines two equal straight lines at right angles to one another.

*17. Three given straight lines meet at a point; draw another straight line so that the two portions of it intercepted between the given lines may be equal to one another.

18. From a given point draw three straight lines of given lengths, so that their extremities may be in the same straight line, and intercept equal distances on that line.

[See Fig. to 1. 16.] 19. Use the properties of the equilateral triangle to trisect a given finite straight line.

20. In a given triangle inscribe a rhombus, having one of its angles coinciding with an angle of the triangle.

VI.

ON THE CONCURRENCE OF STRAIGHT LINES IN A

TRIANGLE.

DEFINITIONS. (i) Three or more straight lines are said to be concurrent when they meet in one point.

(ii) Three or more points are said to be collinear when they lie upon one straight line.

Obs. We here give some propositions relating to the concurrence of certain groups of straight lines drawn in a triangle : the importance of these theorems will be more fully appreciated when the student is familiar with Books III, and iv.

1. The perpendiculars drawn to the sides of a triangle from their middle points are concurrent.

Let ABC be a A, and X, Y, Z the middle points of its sides.

Then shall the perps drawn to the sides from X, Y, Z be concurrent.

From Z and Y draw perps to AB, AC; these perps, since they cannot be parallel, will meet at some point O.

Ax. 12.
Join OX.

B

х

Нур. .

It is required to prove that OX is perp. to BC.

Join OA, OB, OC.
In the 48 OYA, OYC,
YANYC,

Нур. .
Because

and OY is common to both ;
also the LOYA=the L OYC, being rt. 28;
:: OA=OC.

I. 4.
Similarly, from the As OZA, OZB,
it may be proved that OA=OB.
Hence OA, OB, OC are all equal.
Again, in the As OXB, OXC,

BX=CX,
Because and XO is common to both;
also OB=OC:

Proved. : the L OXB=the LOXC;

I. 8. but these are adjacent 28 ; :: they are rt. 48 ;

Def. 10. that is, OX is perp. to BC. Hence the three perps OX, OY, OZ meet at the point O.

Q.E.D. 2. The bisectors of the angles of a triangle are concurrent.

Let ABC be a A. Bisect the 28 ABC, BCA, by sti aight lines which must meet at some point 0.

Ax. 12.

R
Join AO.
It is required to prove that AO bisects the

L BÂC.
From O draw OP, OQ, OR

perp. to the sides of the A.

B

Р
Then in the As OBP, OBR,
the L OBP=the L OBR,

Constr.
Because and the L OPB=the L ORB, being rt. 28,

and OB is common ; :: OP=OR.

I. 26

Similarly from the A: OCP, OCQ, it may be shewn that OP=OQ, :: OP, OQ, OR are all equal.

R

1Q Again in the AS ORA, OQA,

the L: ORA, OQA are rt. 28, Because

and the hypotenuse OA is
common,

B

Р
also OR=OQ; Proved.
:: the L RAO=the L QAO. Ex. 12, p. 99.

That is, AO is the bisector of the L BAC.
Hence the bisectors of the three _8 meet at the point O.

Q.E.D.

3. The bisectors of two exterior angles of a triangle and the bisector of the third angle are concurrent.

Let ABC be a A, of which the sides AB, AC are produced to any points D and E.

Then shall the bisectors of the 48 DBC, ECB, BAC be concurrent.

Bisect the 28 DBC, ECB by straight lines which must meet at some point O. Ax. 12.

В.

P Р
Join AO.
It is required to prove that AO bisects the

R angle BAC.

From O draw OP, OQ, OR perp. to the sides of the A.

D

E Then in the As OBP, OBR, the L OBP=the L OBR,

Constr.
Because { also the L OPB=the L ORB, being rt. 2o,

and OB is common;
:: OP=OR.

1. 26.
Similarly from the A- OCP, OCQ,
it may be shewn that OP=OQ:

OP, OQ, OR are all equal.
Again in the As ORA, OQA,

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..

the Ls ORA, OQA are rt. 2*, Because and the hypotenuse OA is common, also OR=OQ;

Proved. : the L RAO=the L QAO. Ex. 12, p. 99. That is, AO is the bisector of the L BAC. :: the bisectors of the two exterior _8 DBC, ECB, and of the interior L BAC meet at the point O.

Q.E.D.

{

4. The medians of a triangle are concurrent.

Let ABC be a A.
Then shall its three medians be concurrent.
Let BY and CZ be two of its medians, and
let them intersect at O.
Join AO,

Z
and produce it to meet BC in X.
It is required to shew that AX is the remaining
median of the A.
Through C draw CK parallel to BY :

B

X
produce AX to meet CK at K.
Join BK.

K In the A AKC, because Y is the middle point of AC, and YO is parallel to CK,

:: O is the middle point of AK. Ex. 1, p. 104.

Again in the A ABK,
since Z and O are the middle points of AB, AK,

.. ZO is parallel to BK, Ex. 2, p. 104. that is, OC is parallel to BK:

the figure BKCO is a parm.
But the diagonals of a parm bisect one another, Ex. 5, p. 70.

X is the middle point of BC.
That is, AX is a median of the A.
Hence the three medians meet at the point O. Q.E.D.

COROLLARY. The three medians of a triangle cut one another at a point of trisection, the greater segment in each being towards the angular point. For in the above figure it has been proved that

AO=OK,
also that OX is half of OK ;

:: OX is half of OA:
that is, OX is one third of AX.
Similarly OY is one third of BY,
and OZ is one third of CZ.

Q.E.D.

By means of this Corollary it may be shewn that in any triangle the shorter median bisects the greater side.

[The point of intersection of the three medians of a triangle is called the centroid. It is shewn in Mechanics that a thin triangular plate will balance in any position about this point : therefore the centroid of a triangle is also its centre of gravity.]

H.S.E.

H

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

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Let ABC be a A, and AD, BE, CF the three perpå drawn from the vertices to the opposite sides.

Then shall the perps AD, BE, CF be concurrent. Through A, B, and C draw straight lines MN, NL, LM parallel to the opposite sides of the A. Then the figure BAMC is a par.

Def. 36. . AB=MC.

I. 34.
Also the figure

BACL is a parm.
AB=LC,

LC=CM: that is, C is the middle point of LM. So also A and B are the middle points of MN and NL. Hence AD, BE, CF are the perps to the sides of the A LMN from their middle points.

Ex. 3, p. 60. But these perpå meet in a point: Ex. 1, p. 111. that is, the perps drawn from the vertices of the A ABC to the opposite sides meet in a point.

Q.E.D. [For another proof see Theorems and Examples on Book 111.)

DEFINITIONS.

(i) The intersection of the perpendiculars drawn from the vertices of a triangle to the opposite sides is called its orthocentre.

(ii) The triangle formed by joining the feet of the perpendiculars is called the pedal triangle.

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