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sides ba and ac, but (2) the angle bdc will be greater than

the angle bac.

(1) For proof continue bd to f.

Then ba+ af greater than bf (being two sides of triangle baf. See Sect. 2). Add to each fc, then ba + ac greater than bf + fc.

Also dffc greater than dc (Sect. 2), and adding to each bd, bf + fc greater than bd + dc, and ba and ac having been shown to be greater than bf+fc, much more are they greater than bd + dc.

(2) For proof, angle bdc as exterior angle of triangle dfc is greater than interior and opposed angle cfd. (Sect. 3.) Also angle cfd as exterior angle of triangle fab is greater than internal and opposed angle baf. (Sect. 3.) Therefore, a fortiori, the angle angle bac.

bdc is greater than

6b. If, however, one of the lines be drawn from the base, but not from an end of it, the interior lines may exceed in length the exterior lines.

This is a modification of Proposition 21 in the Notes to Playfair's Geometry.

In Fig. D 36, let abc be a triangle with a right angle at a, and from any point (d) in the base draw the line dc. Set off df=ac, and bisect cf at point g, and draw line gb. Then lines dg and gb will together be greater than ac and cb.

For cg (or its equal gf) and gb are greater than cb (D 2), and gd greater than ac (which = df), .. dg and gb together are greater than ac and cb.

7. If two triangles have two sides equal, but the included

angle is greater in one than in the other, the base of that having the greater angle will be the greater. (Euclid 1, 24.)

In Fig. D 4, let the triangles bac and bad have the side ba common and sides ac and ad equal, but the included angle bac greater than the included angle bad.

Then base be will be greater than bd.

[Note. In the figure the sides ac and ad are taken as longer than the common side ab, as otherwise, without affecting the accuracy of the proposition, three different cases will arise.]

Draw the line dc, completing the triangles acd and bcd. Then in acd sides ac and ad being equal, angle acd = angle adc (C 15).

And in triangle bcd angle bdc is greater than angle adc, and a fortiori greater than angle bcd (part of angle acd).

Then side bc (base of triangle bac) opposed to greater angle will be greater than side bd (base of triangle bad) opposed to the less angle bcd.

Another and more simple demonstration is given in Sect. 12.

8. If two triangles have two sides equal, but different lengths of base, the larger base will have the larger angle included between the equal sides. (Euclid 1, 25.)

This is the converse of Sect. 7.

In Fig. D 4 let base bc be greater than base bd.

Then the included angle bac will be greater than the included angle bad.

The figure shows obviously that such is the case, but for proof.

If not greater, the angle must be equal or less.

If equal, the side ad would cover side ac, which it does not.

If less, be would be less than bd, but it is greater.
Being therefore neither equal nor less it must be greater.

9. In triangles having the same base, and the summits touching a given right line, the sum of the two other sides will be the least where those sides make an equal angle with the given line. (Bezout, Reynaud's Notes, Sect. 10.)

In Fig. ▷ 5 let ab be the given base, xy the given line, and acb, adb, and afb three given triangles, of which in acb the sides ac and be make an equal angle with the given line; then the sum of the sides ac and be will be less than the sum of the sides ad and db, and than the sum of the sides af and fb in the two triangles in which those sides do not make an equal angle with the given line.

For the demonstration. From point a and perpendicular to the line xy draw agh, making gh = ga, and draw lines ch, dh, and fh.

Then gh being equal to ga and the line xy perpendicular (by construction), the obliques ad and hd are equal, as also the obliques ac and he and af and hƒ (C 30, 3).

Consequently bd and bh bd and da, the two sides of triangle adb; bc and ch = the two sides of triangle acb; and bƒ and fh the two sides of triangle afb.

But bch forms base of the triangles of which bdh and bfh form respectively two other sides, and it is consequently less than the two other sides in each case (Sect. 2).

And the equal of bch being bca will therefore be less than adb, the equal of bdh, or than afb, the equal of bfh.

10. Although the sides will be less, as in Sect. 9, the area

will be greater or less, as the perpendicular height of the triangle is greater or less (Bezout, Reynaud's Notes, Sects. 11, 12).

In Fig. D 6, let there be three triangles, having the same base ab and the summits touching the right line xy, the triangle acb having the two sides ac and cb making equal angles with the given line, and the triangles adb and afb making unequal angles, but the altitudes being different as in the figure.

Then the area of triangle adb, having the greater altitude, will be greater than the triangle acb, whilst the area of the triangle afb, with less altitude, will be less.

Draw the parallels 1, 2, and 3, passing through the summits of the several triangles, and draw the dotted lines ai and aj.

Then triangles afb and aib will be equal in area, having the same base and being between the same parallels (C 17).

But triangle aib is less than triangle acb, having the greater altitude, by the triangle aji.

Comparing the triangle acb with the triangle adb, having greater altitude; the triangles acb and ajb will be equal in area (C 17), and triangle adb will exceed that triangle by the triangle adj.

11. When the base forms a chord, and the summits are bounded by an arc, the sides, having equal angles at the base and equal lengths, will be the greatest (Bezout, Reynaud, Sect. 10).

In Fig. 7 D it will be seen that the triangles bounded by the arc acdb will differ materially from those bounded by a right line mn touching the summits.

The demonstration of the sides ac and cb being the

I

greatest is given in Sect. 27 of Div. E post, but it is thought desirable to draw attention to the difference between the summits of triangles on the same base touching a line or an arc of a circle.

12. Of triangles having one side common and which cut one another, the sum of the two sides which cut will be greater than that of the two sides which do not cut. (Rouche and Comberousse, Sect. 32, Fig. 24.)

In Fig. D 4, let the two triangles acb and adb have the side ab common and the two sides ad and be cut each other

at n.

Then the cutting sides ad and bc will be together greater than the sides bd and ac which do not cut.

The parts bn and nd of the cutting sides are together greater than the side bd, and the remaining parts an and nc are together greater than ac, the parts in each case being two sides of a triangle, of which bd in the one case and ac in the other will form the third side. (Sect. 2.)

Added together the sum of the two cutting sides bc and ad will be greater than that of sides bd and ac.

13. In a triangle, the third side of which passes through a given point, the triangle in which such side is bisected at that point will contain the least area. (Hutton's Math. V. 3.)

In Fig. D 8, Case 1, let abc be an isosceles triangle having the third side pass through and be bisected at point d.

Then that triangle will have a less area than a triangle of which the third side fg passes through but is not bisected by point d.

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