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PROPOSITION VII. THEOREM.

If two triangles have one angle of the one equal to one angle of the other, and the sides about a second angle in each proportionals; then, if the third angles in each be both acute, both obtuse, or if one of them be a right angle, the triangles must be equiangular to one another, and must have those angles equal, about which the sides are proportionals.

AA

In the As ABC, DEF, let

BAC= 2 EDF, and let AB be to BC as DE is to EF,

and let s ACB, DFE be both acute, both obtuse, or let one of them be a right angle.

Then must As ABC, DEF be equiangular to one another, having ABC= ▲ DEF, and ▲ ACB= L DFE.

For if ABC be not ▲ DEF, let one of them, as ▲ ABC, be greater than the other, and make 4 ABG=▲ DEF, I. 23.

and let BG meet AC in G.

Then *.* 2 BAG= ▲ EDF, and ▲ ABG= 2 DEF,

..▲ ABG is equiangular to ▲ DEF,

and.. AB is to BG as DE is to EF.

But AB is to BC as DE is to EF,

.. AB is to BG as AB is to BC,

and .. BG=BC,

and.. BCG= 4 BGC.

I. 32.

VI.4.

Hyp.

V. 5.

V. 8.

I. A.

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..4 BCG is obtuse, which is contrary to the hypothesis.

Next, let ▲ ACB and ▲ DFE be both obtuse,

I. 13.

then AGB is obtuse, and .. 4 BGC is acute; .. BCG is acute, which is contrary to the hypothesis. Lastly, let one of the third s ACB, DFE be a right ▲ . If ACB be a rt.,

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and .. 4s BCG, BGC together=two rt. 48,

which is impossible.

I. 17.

Hence ABC is not greater than ▲ DEF.

So also we might shew that 4 DEF is not greater than

L ABC.

..4 ABC = L DEF,

and .. ACB = 4 DFE.

I. 32.

Q. E. D.

N.B.-This Proposition is an extension of Proposition E of Book I. p. 42.

Note. We have made a slight change in Euclid's arrangement of the four Propositions that follow, because Eucl. vi. 8 is closely connected with the proof of Eucl. vi. 13.

PROPOSITION VIII. PROBLEM.

(Eucl. vi. 9.)

From a given straight line to cut off any submultiple.

B

E

Let AB be the given st. line.

It is required to shew how to cut off any submultiple from AB. From A draw AC making any angle with AB.

In AC take any pt. D, and make AC the same multiple of AD that AB is of the submultiple to be cut off from it.

Join BC, and draw DE || to BC.

Then. ED is || to BC,

.. CD is to DA as BE is to EA,

and.. CA is to DA as BA is to EA.

I. 31.

VI. 2.

V. 16.

.. EA is the same submultiple of BA that DA is of CA.

V. 19.

Hence from AB the submultiple required is cut off.

Q. E. F.

Ex. 1. Cut off one-seventh of a given straight line.
Ex. 2. Cut off two-fifths of a given straight line.

Note. This Proposition is a particular case of Proposition IX.

PROPOSITION IX. PROBLEM. (Eucl. vI. 10.)

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

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Let AB be the st. line given to be divided, and AC the divided st. line.

It is required to divide AB similarly to AC.

Let AC be divided in the pts. D, E.

Place AB, AC so as to contain any angle.

Join BC, and through D, E draw DF, EG || to BC.
Through D draw DHK || to AB.

I. 31.

I. 31.

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Ex. 1. Produce a given straight line, so that the whole produced line shall be to the produced part in a given ratio.

Ex. 2. On a given base describe a triangle, with a given vertical angle and its sides in a given ratio.

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 XI.

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.

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