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The practical application of the foregoing hints is illustrated by the following examples.

1. Construct an isosceles triangle having given the base, and the sum of one of the equal sides and the perpendicular drawn from the vertex to the base.

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Let AB be the given base, and K the sum of one side and the perpendicular drawn from the vertex to the base. ANALYSIS. Suppose ABC to be the required triangle.

From C draw CX perpendicular to AB:
then AB is bisected at X.

1. 26. Now if we produce XC to H, making XH equal to K,

it follows that CH=CA;

and if AH is joined,
we notice that the angle CÂH=the angle CHA.

I. 5. Now the straight lines XH and AH can be drawn before the position of C is known ;

Hence we have the following construction, which we arrange synthetically. SYNTHESIS.

Bisect AB at X: from X draw XH perpendicular to AB, making XH equal to K.

Join AH.
At the point A in HA, make the angle HAC equal to the angle AHX.

Join CB.
Then ACB shall be the triangle required.
First the triangle is isosceles, for AC=BC. 1. 4.
Again, since the angle HAC=the angle AHC, Constr.
:: HC=AC.

1. 6.
To each add CX;
then the sum of AC, CX=the sum of HC, CX

=HX. That is, the sum of AC, CX=K,

Q. E. F,

2. To divide a given straight line so that the square on one part may be double of the square on the other.

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Let AB be the given straight line.

ANALYSIS. Suppose AB to be divided as required at X: that is, suppose the square on AX to be double of the square on XB.

Now we remember that in an isosceles right-angled triangle, the square on the hypotenuse is double of the square on either of the equal sides.

This suggests to us to draw BC perpendicular to AB, to make BC equal to BX, and to join XC. Then the square on XC is double of the square on XB; I. 47.

:: XC=AX.
Hence when we join AC, we notice that
the angle XÁC=the angle XCA.

1. 5. Thus the exterior angle CXB is double of the angle XAC.

1. 32. But the angle CXB is half of a right angle;

1. 32. : the angle XAC is one-fourth of a right angle.

This supplies the clue to the following construction :--
SYNTHESIS. From B draw BD perpendicular to AB;

and from A draw AC, making BAC one-fourth of a right angle. From C, the intersection of AC and BD, draw CX, making the angle ACX equal to the angle BAC.

I. 23.
Then AB shall be divided as required at X.
For since the angle XCA=the angle XAC,
: XA=XC.

I. 6. And because the angle BXC=the sum of the angles BAC, ACX, 1. 32.

:. the angle BXC is half a right angle.

And the angle at B is a right angle ;
:. the angle BCX is half a right angle ;
:. the angle BXC=the angle BCX;

BX=BC. Hence the square on XC is double of the square on XB : I. 47. that is, the square on AX is double of the square on XB. Q. E. F.

1. 32.

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I.

ON THE IDENTICAL EQUALITY OF TRIANGLES.

See Propositions 4, 8, 26.

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1. If in a triangle the perpendicular from the vertex on the base bisects the base, then the triangle is isosceles.

2. If the bisector of the vertical angle of a triangle is also perpendicular to the base, the triangle is isosceles.

3. If the bisector of the vertical angle of a triangle also bisects the base, the triangle is isosceles.

[Produce the bisector, and complete the construction after the manner of 1. 16.]

4. If in a triangle a pair of straight lines drawn from the ex. tremities of the base, making equal angles with the remaining sides, are equal, the triangle is isosceles.

5. If in a triangle the perpendiculars drawn from the extremities of the base to the opposite sides are equal, the triangle is isosceles.

6. Two triangles ABC, ABD on the same base AB, and on opposite sides of it, are such that AC is equal to AD, and BC is equal to BD: shew that the line joining the points C and D is perpendicular to AB.

7. If from the extremities of the base of an isosceles triangle perpendiculars are drawn to the opposite sides, shew that the straight line joining the vertex to the intersection of these per. pendiculars bisects the vertical angle.

8. ABC is a triangle in which the vertical angle BAC is bisected by the straight line AX : from B draw BD perpendicular to AX, and produce it to meet AC, or AC produced, in E; then shew that BD is equal to DE.

9. In a quadrilateral ABCD, AB is equal to AD, and BC is equal to DC: shew that the diagonal AC bisects each of the angles which it joins.

10. In a quadrilateral ABCD the opposite sides AD, BC are eqnal, and also the diagonals AC, BD are equal: if AC and BD intersect at K, shew that each of the triangles ÅKB, DKC is isosceles.

11. If one angle of a triangle be equal to the sum of the other two, the greatest side double of the distance of its middle point from the opposite angle.

12. Two right-angled triangles which have their hypotenuses equal, and one side of one equal to one side of the other, are equal in all respects.

A

B

E

F

Let ABC, DEF be two $ right-angled at B and E, having AC equal to DF, and AB equal to DĚ.

Then shall the A ABC be equal to the A DEF in all respects.

For apply the A ABC to the A DEF, so that AB may coincide with the equal line DE, and C may fall on the side of DE remote from F. Let C' be the point on which C falls. Then DEC' represents the A ABC in its new position. Now each of the Ls DEF, DEC' is a rt. 2; Нур. .

.:. EF and EC' are in one st. line. Then in the A C'DF, because DF=DC' (i.e. AC), Hyp.

.: the LDFC'=the L DC'F. Hence in the two As DEF, DEC',

the L DEF=the L DEC', being rt. 28; Because and the LDFE=the L DC'E;

Proved. also the side DE is common to both; :: the As DEF, DEC' are equal in all respects;

1. 26. that is, the As DEF, ABC are equal in all respects. Q.E.D.

1. 14.

1. 5.

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Alternative Proof. Since the L ABC is a rt. angle; :: the sq. on AC=the sqq. on AB, BC.

I. 47. Similarly, the sq. on DF = the sqq. on DE, EF ; I. 47. But the sq. on AC=the sq. on DF, since AC=DF;

: the sqq. on AB, BC=the sqq. on DE, EF. And of these, the sq. on AB=the sq. on DE, since AB=DE; :: the sq. on BC=the sq. on EF ;

Ax. 3. .: BC=EF. Hence the three sides of the A ABC are respectively equal to the three sides of the A DEF;

:: the A ABC=the A DEF in all respects. I. S.

13. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles opposite to one pair of equal sides equal, then the angles opposite to the other pair of equal sides shall be either equal or supplementary, and in the former case the triangles shall be equal in all respects.

E

B с
Fig. 1.

E F
Fig. 2.

F'
Fig. 3.

Let ABC, DEF be two triangles, in which

the side AB=the side DE,

the side AC=the side DF,

and the LABC=the LDEF. Then shall the L8 ACB, DFE be either equal (as in Figs. 1 and 2) or supplementary (as in Figs. 1 and 3); and in the former case the triangles shall be equal in all respects.

If the LBAC=the LEDF. [Figs. 1 and 2.] then the LACB=the LDFE, and the triangles are equal in all respects.

1. 4. But if the LBAC be not equal to the LEDF, [Figs. 1 and 3.]

let the LEDF be greater than the 2 BAC.
At D in ED make the L EDF' equal to the 2 BAC.
Then the As BAC, EDF' are equal in all respects.

I. 26.
:. AC=DF';
but AC=DF;

Нур. . .: DF=DF', .: the angle DFF'=the 2 DF'F.

1. 5. But the Ls DF'F, DF'E are supplementary, I. 13.

the _s DFF', DF'E are supplementary : that is, the 48 DFE, ACB are supplementary.

Q.E.D.

COROLLARIES. Three cases of this theorem deserve special attention.

It has been proved that if the angles ACB, DFE are not supple. mentary they are equal :

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