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PROPOSITION X. PROBLEM. (Eucl. vi. 11.)

To find a THIRD proportional to two given straight lines.

B

D

Let AB and AC be the given st. lines.

E

It is required to find a third proportional to AB, AC. Place AB, AC so as to contain any angle.

Produce AB, AC to D and E, making BD=AC. I. 3. Join BC, and through D draw DE || to BC.

Then

BC is || to DE,

.. AB is to BD as AC is to CE,

and.. AB is to AC as AC is to CE. Thus CE is a third proportional to AB and AC.

I. 31.

VI. 2.

V. 6.

Q. E. F.

NOTE. This Proposition is a particular case of Proposition XL

DEF. II. When three magnitudes are proportionals, the first is said to have to the third the duplicate ratio of that, which it has to the second.

Thus here AB has to CE the duplicate ratio of AB to AC.

DEF. III. When three magnitudes are proportionals, the first is said to have to the third the ratio compounded of the ratio, which the first has to the second, and of the ratio, which the second has to the third.

Thus here AB has to CE the ratio compounded of the ratios of AB to AC and AC to CF.

PROPOSITION XI. THEOREM. (Eucl. vi. 12.)

To find a FOURTH proportional to three given straight

lines.

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Let A, B, C be the three given st. lines.

It is required to find a fourth proportional to A, B, C.

Take DE, DF, two st. lines making an EDF, and in these make DG=A, GE=B, and DH=C,

and through E draw EF || to GH.

[blocks in formation]

I. 3.

I. 31.

VI.2.

V. 6.

Thus HF is a fourth proportional to A, B, C.

Q. E. F.

Ex. ABC is a triangle inscribed in a circle, and BD is drawn to meet the tangent to the circle at A in D, at an angle ABD equal to the angle ABC. Show that AC is a fourth proportional to the lines BD, DA, AB.

PROPOSITION XII. THEOREM. (Eucl. vi. 8.)

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

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Let ABC be a right-angled ▲, having ▲ BAC a rt. 4, and from A let AD be drawn 1 to BC.

Then must as DBA, DAC be similar to ▲ ABC, and to each other.

Forrt.

BDA=rt. ▲ BAC, and ▲ ABD= ▲ CBA,

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.. ▲ DBA is equiangular, and .. similar to ▲ ABC. VI. 4. In the same way it may be shown

that ADAC is equiangular, and .. similar to ▲ ABC. Hence ▲ DBA is similar to ▲ DAC.

Q. E. D.

COR. I. DA is a mean proportional between BD and DC, For BD is to DA as DA is to DC.

VI. 4.

COR. II. BA is a mean proportional between BC and BD, For BC is to BA as BA is to BD.

VI. 4.

COR. III. CA is a mean proportional between BC and CD, For BC is to CA as CA is to CD. VI. 4.

Q. E. D.

Ex. B is a fixed point in the circumference of a circle, whose centre is C; PA is a tangent at any point P, meeting CB produced in A, and PD is drawn perpendicularly to CB. Prove that the line bisecting the angle APD always passes through B.

PROPOSITION XIII.

PROBLEM.

To find a MEAN proportional between two given straight

lines.

B

Let AB and BC be the two given st. lines.

It is required to find a mean proportional between AB and BC.

Place AB and BC so as to make one st. line AC, and on AC describe the semicircle ADC.

From B draw BD1 to AC, and join AD, CD.

[blocks in formation]

I. 11. III. 31.

.. DB is a mean proportional between AB and BC.

VI. 12, COR. 1.

Q. E. F.

Ex. 1. Produce a given straight line, so that the given line may be a mean proportional between the whole line and the part produced.

Ex 2 Shew that either of the sides of an isosceles triangle is a mean proportional between the base and the half of the segment of the base, produced if necessary, which is cut off by a straight line, drawn from the vertex, at right angles to the equal side.

Ex. 3. Shew that the diameter of a circle is a mean proportional between the sides of an equilateral triangle and a hexagon, described about the circle.

Ex. 4. From a point A, outside a circle, a line is drawn, cutting the circle in B and C. Find a mean proportionalbetween AB and AC.

DEF. IV. Two figures are said to have their sides about two of their angles reciprocally proportional, when, of the four terms of the proportion, the first antecedent and the second consequent are sides of one figure, and the second antecedent and first consequent are sides of the other figure.

Thus, in the diagram on the opposite page, the figures AB and BC have their sides about the angles at B reciprocally proportional, the order of the proportion being

DB is to BE as GB is to BF.

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