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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 perpå to AB, AC; these perp”, since they cannot be parallel, will meet at some point O. Ax. 12. Join OX.

B

х

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

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

Нур.
Because

and OY is common to both;
also the LOYA=the L OYC, being rt. _$ ;
:: 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,

Hyp.
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. 28 ;

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 _8 ABC, BCA, by straight lines which must meet at some point 0.

Ax. 12.

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

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

B

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

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

and OB is common; . OP=OR.

1. 26.

R

Similarly from the As OCP, OCQ, it may be shewn that OP=OQ, :: OP, OQ, OR are all equal. Again in the As ORA, OQA,

the Ls ORA, OQA are rt. LS, Because

and the hypotenuse OA is
common,

B

P
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 _s meet at the point O.

Q.E.D.

P

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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. Join AO.

В.

С It is required to prove that AO bisects the

RA angle BAC.

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

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

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

and OB is common;

OP=OR.
Similarly from the As OCP, OCQ,
it may be shewn that OP=OQ:
:: OP, OQ, OR are all equal.
Again in the A$ ORA, OQA,

the L8 ORA, OQA are rt. 28,
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 _s DBC, ECB, and of the interior L BAC meet at the point O.

Q.E.D.

..

I, 26.

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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 :

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

ÃO=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.SE,

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 perps 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 parm.

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 ALMN from their middle points.

Ex. 3, p. 60. But these perps meet in a point: Ex. 1, p. 1ll. 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.

VII.

ON THE CONSTRUCTION OF TRIANGLES WITH GIVEN

PARTS.

a

Obs. No general rules can be laid down for the solution of problems in this section ; but in a few typical cases we give constructions, which the student will find little difficulty in adapting to other questions of the same class.

1. Construct a right-angled triangle, having given the hypotenuse and the sum of the remaining sides.

[It is required to construct a rt.angled A, having its hypotenuse equal to the given straight line K, and the sum of its remaining sides equal to AB. From A draw AE making with BA

K an equal to half a rt. L. From centre B, with radius equal to K, de. scribe a circle cutting AE in the points C, C'.

А. D D B From C and C draw perps CD, C'D' to AB; and join CB, C'B. Then either of the A$ CÔB, C'D'B will satisfy the given conditions.

NOTE. If the given hypotenuse K be greater than the perpendicular drawn from B to AE, there will be two solutions. If the line K be equal to this perpendicular, there will be one solution ; but if

1; less, the problem is impossible.]

2. Construct a right-angled triangle, having given the hypotenuse and the difference of the remaining sides.

3. Construct an isosceles right-angled triangle, having given the sum of the hypotenuse and one side.

4. Construct a triangle, having given the perimeter and the angles at the base.

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[Let AB be the perimeter of the required A, and X and Y the 28 at the base.

From A draw AP, making the 2 BAP equal to half the L X.
From B draw BP, making the L ABP equal to half the L Y.
From P draw PQ, making the L APQ equal to the 2 BAP.
From P draw PR, making the L BPR equal to the 2 ABP.

Then shall PQR be the required A.]

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