This unravelling of the conditions of a proposition in order to trace it back to some earlier principle on which it depends, is called geometrical analysis: it is the natural way of attacking most exercises of a more difficult type, and it is especially adapted to the solution of problems.

These directions are so general that they cannot be said to amount to a method: all that can be claimed for Geometrical Analysis is that it furnishes a mode of searching for a suggestion, and its success will necessarily depend on the skill and ingenuity with which it is employed: these may be expected to come with experience, but a thorough grasp of the chief Propositions of Euclid is essential to attaining them.

The practical application of these 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.

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.

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 CAH the angle CHA.

I. 26.

1. 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.

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; and join CB.

Then ACB shall be the triangle required.

First the triangle is isosceles, for AC = BC.
Again, since the angle HAC the angle AHC,

HC=AC.

To each add CX;

then the sum of AC, CX= the sum of HC, CX

=HX.

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

I. 4. Constr.

1. 6.

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.

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, and to make BC equal to BX.

Join XC.

Then the square on XC is double of the square on XB,

XC=AX.

And when we join AC, we notice that

the angle XAC= the angle XCA.

Hence the exterior angle CXB is double of the angle XAC.

But the angle CXB is half of a right angle:

the angle XAC is one-fourth of a right angle.

This supplies the clue to the following construction:

I. 47.

1. 5.

I. 32.

I. 32.

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.

Then AB shall be divided as required at X.

For since the angle XCA = the angle XAC,

XA=XC.

I. 23.

1. 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;

therefore the angle BCX is half a right angle;
therefore the angle BXC = the angle BCX;

BX = BC.

1. 32.

Hence the square on XC is double of the square on XB: 1. 47. that is, the square on AX is double of the square on XB. Q.E.F.

I. ON THE IDENTICAL EQUALITY OF TRIANGLES.

See Propositions 4, 8, 26.

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 extremities of the base, making equal angles with the 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 perpendiculars 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 equal, and also the diagonals AC, BD are equal: if AC and BD intersect at K, shew that each of the triangles AKB, DKC is isosceles.

11. If one angle of a triangle be equal to the sum of the other two, the greatest side is 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 identically equal.

A

D

B

E

F

S

Let ABC, DEF be two As right-angled at B and E, having AC equal to DF, and AB equal to DE:

then shall the As be identically equal.

For apply the A ABC to the A DEF, so that A may fall on D, and AB along DE; and so that C may fall on the side of DE remote from F.

Let C' be the point on which C falls.

Then since AB=DE,

.. B must fall on E;

so that DEC' represents the ▲ ABC in its new position.

Now each of the S DEF, DEC' is a rt. L;

.. EF and EC' are in one st. line.

Then in the ▲ C'DF,
because DF DC',

.. the DFC' the DC'F.

Hence in the two AS DEF, DEC',

the DEF the DEC', being rt. LS;

also the side DE is common to both;

.. the AS DEF, DEC' are equal in all respects; that is, the AS DEF, ABC are equal in all respects.

Hyp.

I. 14.

1. 5.

Proved.

I. 26.

Q.E.D.

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 are either equal or supplementary, and in the former case the triangles are equal in all respects.

444

B

CE

FE

F'

F

Let ABC, DEF be two triangles, having the side AB equal to the side DE, the side AC equal to the side DF, and the L ABC equal to the L DEF; then shall the LS ACB, DFE be either equal or supplementary, and in the former case the triangles shall be equal in all respects.

If the L BAC the L EDF,

=

then the triangles are equal in all respects.

I. 4.

But if the BAC be not equal to the EDF, one of them must be

the greater.

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

..the

.. AC=DF' ; but ACDF;

.. DF=DF',

DFF' the L DF'F.

But the LS DF'F, DF'E are supplementary, ..the DFF', DF'E are supplementary: that is, the LS DFE, ACB are supplementary.

Three cases of this theorem deserve special attention.

I. 26.

Hyp.

I. 5.

I. 13.

Q. E. D.

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

And we know that of angles which are supplementary and unequal, one must be acute and the other obtuse.