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To find a mean proportional between two given straight lines.


Let it be required to find a mean proportional between A and BC.

Produce BC making the produced part CHA.

On BH describe a semicircle and draw CX 1 to BH. Then CX shall be a mean proportional between A and BC.

Join BX, XH.

Then As BCX, XCH are equiangular ;

.. CH: CX as CX : BC;

(VI. 4)

.. CX is a mean proportional between CH and BC;

i.e. between A and BC.


Divide a straight line into two parts so that one of the parts shall be a mean proportional between the whole line and the other part.

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Let PQ be the given straight line.
Divide it in X so that the rect. PQ, QX may be

on PX.

Then PQ: PX as PX : QX.

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PQ is said to be divided in extreme and mean ratio in X.




Similar triangles are to one another as the squares on their

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Let ABC, AHK be similar as;

Then ▲ ABC: ▲ AHK as square on AB : square on AH the corresponding side of ▲ AHK.

Let the ▲ ABC be applied to the ▲ AHK so as to have the sides AB, AC on the corresponding sides AH, AK; and... having the side BC || to HK.

On AB, AH describe the squares AE, AG.

Produce DE to meet GH in X; and join CH.

Then a ABC: ▲ AHC as AB : AH, and .. as AE: AX;

(v. 1)

also ▲ AHC ▲ AHK as AC : AK;

(v. I)

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Similar polygons are to one another as the squares on their corresponding sides.

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Let ABCDF, GHKMT be similar polygons.

Then shall they bear to one another the same ratio as the squares on their corresponding sides AB, GH.

Divide the polygons into triangles by straight lines drawn from В and H.

Then . FA: AB as TG: GH, and FAB = L TGH; .. As FAB, TGH are similar; (VI. 4) and are to each other as squares on FB, TH. (v1. 7) Again, ·.· ↳ DFA LMTG and ▲ BFA = L HTG; .. L DFB = L MTH;


also DF: FA as MT: TG;
and FA : FB as TG: TH;

:. DF: FB as MT: TH;

(V. 12) (VI. 4)

.. As DFB, MTH are similar; and are to each other as squares on FB, TH, (v1. 7) and are ... in the same ratio as ▲s FAB, TGH. So each of the As in the polygon ABCDF: the corresponding ▲ in the polygon GHKMT in the same ratio.

... all the ▲s in the polygon ABCDF bear to all the As in polygon GHKMT the same ratio. (V. II) That is, the area of polygon ABCDF: area of polygon GHKMT as ▲ FAB: ▲ TGH,

(VI. 7)

and .. as square on AB : square on GH. COR. Each of the sides of the one polygon is to the corresponding side of the other as AB: GH

... also the perimeter of the one is to the perimeter of the other in the same ratio.

(V. II)


If four straight lines are proportional, the squares on these lines are proportional.

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Let AB, AH, AC, AK be cut off from the arms of any A the given st. lines and ... proportional.


Then shall the squares on AB, AH, AC, AK be proportional also.

Join BC, HK.

Then BC is to HK,

and .. ▲s ABC, AHK are similar;

(VI. I)

(I. 21)

.. ▲ ABC : ▲ AHK as square on AB : square on AH,

and also

(VI. 7) as square on AC: square on AK;

(VI. 7)

... sq. on AB : sq. on AH as sq. on AC: sq. on AK. (v. 3)

COR. Hence also, if similar figures are described on AB and AH and similar figures on AC and AK,


since figure on AB similar figure on AH as square on AB : square on AH,

and figure on AC: similar figure on AK as square on AC square on AK, it follows that

fig. on AB sim. fig. on AH as fig. on AC: sim. fig.

on AK

(v. 3)


In right-angled triangles the rectilineal figure described on the hypothenuse is equal to the similar and similarly described figures on the other two sides.

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Let P, Q, R be figures similar and similarly described on the sides of the right-angled AABC.

Then the figure P on the hypothenuse BC is equal to the figures Q and R together.

From A let fall ADL to BC.

Then on BC, AC are the similar as ABC, DAC,
and also the similar figures P, Q;

.. ΔΑΒC : ▲ DAC as P : 2; (VI. 9 Cor.)

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DBA figures Q, R together,

but AABC is = AS DAC, DBA together;

.. Pis Q and R together.


(V. II)

(v. 4)

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