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DEFINITION.

Similar rectilineal figures are those which have their corresponding angles equal and the sides about the equal angles proportional.

PROPOSITION VI.

In a right-angled triangle, if a perpendicular be drawn from the right angle to the base; the triangles on each side of it shall be similar to the whole triangle and to one another.

H

Let ACB be a right-angled ▲ having the right ▲ ACB. From Clet fall CH 1 to AB.

Then shall the as AHC, CHB, ACB be similar to one another.

For the A is common to the two as AHC, ACB;
and right AHC = right ▲ ACB;
.. AS AHC, ACB are equiangular.
Similarly as CHB, ACB are equiangular.
Hence the As AHC, CHB, ACB are equiangular and

... similar to one another.

(1. 25)

(VI. 4)

COR. The perpendicular from the right angle is a mean proportional between the segments of the base.

For since

AHC is equiangular to ▲ CHB, .. AH: HC as HC: HB;

(VI. 4)

.. HC is a mean proportional between AH, HB.

C. G.

12

PROBLEM A.

To divide a straight line similarly to a given divided line.

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Let it be required to divide AB similarly to GH.

Through A draw a straight line AF equal to GH, making any with AB,

and cut off parts AL, LM, MK = GR, RS, ST respectively. Join BF, and through L, M, K draw LX, MY, KZ || to FB.

Then shall AB be divided similarly to GH.

Through L, M, K draw LO, MP, KQ || to AB.

Then as ALX, LMO, MKP, KFQ are equiangular; (1.21)

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and MK: KF as YZ : ZB.

Hence AB is divided similarly to AF, and .. to GH.

PROBLEM B.

To find a fourth proportional to three given straight lines.

H

P Q R

B

Let it be required to find a fourth proportional to the three straight lines P, Q, R.

Draw two st. lines ABX, ACH making any ▲ at A.
Cut off AC = P, CH = Q, and AB = R.
Join CB, and through H draw HX | to CB, cutting

ABX in X.

Then shall BX be a fourth proportional to P, Q, R.

For since HX is || to CB;

.. AC : CH as AB: BX,
. P Q as R BX.

(VI. I)

PROBLEM C.

To find a third proportional to two given straight lines.

P Q

A

B

H

X

Let it be required to find a third proportional to the

two given straight lines P, Q.

Draw two st. lines ABX, ACH making any ▲ at A.

Cut off AC = P, CH = Q, and AB =

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Join CB, and through H draw HX || to CB.
Then shall BX be a third proportional to P, Q.

For since HX is || to CB,

.. AC CH as AB : BX;

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:

Q as Q : BX.

(VI. 1)

PROBLEM D.

To find a mean proportional between two given straight lines.

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

Produce BC making the produced part CH = A.

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.

PROBLEM E.

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.

COMPARISON OF AREAS OF SIMILAR

RECTILINEAL FIGURES.

PROPOSITION VII.

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

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

(v. 1)

also ▲ AHC: ▲ AHK as AC : AK;

(v. 1)

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