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A history of Greek mathematics - Wilbourhall.org

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366 PAPPUS OF ALEXANDRIA<br />

triangle with base less than one <strong>of</strong> the other sides, it is possible<br />

to construct on the base and within the triangle two straight<br />

lines meeting at a point, the sum <strong>of</strong> which is equal to the sum<br />

<strong>of</strong> the other two sides <strong>of</strong> the triangle (Props. 29, 30).<br />

2. In any triangle in which it is possible to construct two<br />

straight lines from the base to one internal point the sum<br />

<strong>of</strong> which is equal to the sum <strong>of</strong> the two sides <strong>of</strong> the triangle,<br />

it is also possible to construct two other such straight lines the<br />

sum <strong>of</strong> which is greater than that sum (Prop. 31).<br />

3. Under the same conditions, if the base is greater than either<br />

<strong>of</strong> the other two sides, two straight lines can be so constructed<br />

from the base to an internal point which are respectively<br />

greater than the other two sides <strong>of</strong> the triangle ; and the lines<br />

may be constructed so as to be respectively equal to the two<br />

sides, if one <strong>of</strong> those two sides is less than the other and each<br />

<strong>of</strong> them is less than the base (Props. 32, 33).<br />

4. The lines may be so constructed that their sum will bear to<br />

the sum <strong>of</strong> the two sides <strong>of</strong> the triangle any ratio less<br />

2 : 1 (Prop. 34).<br />

than<br />

As examples <strong>of</strong> the pro<strong>of</strong>s, we will take the case <strong>of</strong> the<br />

scalene triangle, and prove the first and Part 1<br />

the above propositions for such a triangle.<br />

than<br />

<strong>of</strong> the third <strong>of</strong><br />

In the triangle ABC with base BC let AB he greater<br />

AC.<br />

Take D on BA such that BD = J (BA + AC).<br />

B H L<br />

On DA between D and A take any point E, and draw EF<br />

parallel to BC. Let G be any point on EF; draw GH parallel<br />

to AB and join GC.

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