EXAMPLES.

1. Find the point in the circumference of a circle the sum of its distances from two fixed points in the same circumference shall be a maximum.

[The solution here is identically the same as that in Prop. 72, with the exception that the angle subtended at the circumference of the circle by the two fixed points is not necessarily a right angle: the point required is, then, the bisection of the arc.]

2. Of all triangles that can be inscribed in a circle the equilateral has the maximum perimeter.

[Considering the foregoing, the reasoning is similar to that used in Prop. 87.]

3. Prove, generally, that of all figures inscribable in a circle the regular has the maximum perimeter.

[The same reasoning here, again, applies.]

4. Given, of a triangle, the base and vertical angle: construct it, so that its area shall be a maximum.

[The area is clearly a maximum when the vertex of the triangle is the bisection of the arc of the segment of the circle described upon the given base, and containing an angle equal to the given vertical angle.]

5. Of all triangles that can be inscribed in a given circle, the equilateral has the maximum area.

[This follows from the foregoing, by a reasoning similar to that in Prop. 78.]

6. Of all figures that can be inscribed in a given circle, the regular has the maximum area.

[The same reasoning here, also, applies.]

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MISCELLANEOUS PROPOSITIONS.

PROP. 82. The rectangle under the distances of the middle point of the base of a triangle from the foot of the perpendicular, and the point where the internal bisector of the vertical angle meets the base, is equal to the square of half the difference of the sides.

B

E

A

H

G

Let ABC be any triangle, of which AG, AH are, respectively, the perpendicular, and internal bisector of the vertical angle, meeting the base in G, H. Bisect BC in

GEOMETRICAL EXERCISES FOR BEGINNERS.

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D: then DG. DH is equal to the square of half the difference of the sides.

Cut off AE equal to AC, and join CE, meeting AH in F. Join DF.

Since the angles CAF, EAF are equal by hypothesis, and the sides CA, AE also equal; therefore (I. 4) CF, EF are equal, and AF is perpendicular to CE. Hence (1) DF is parallel to BE, and equal to half of it, that is, DF is half the difference of the sides. Again, the angles DFH, BAH are equal (I. 29); and the angles BAH, CAF are equal (hyp.); but the angles CAF, HGF are equal (17), since a circle will circumscribe the quadrilateral AFGC (the angles AFC, AGC being right); therefore the angles DFH, HGF are equal. Hence (31) DG. DH = DF2, that is, the square of half the difference of the sides.

PROP. 83. The rectangle under the distances of the middle point of the base of a triangle from the foot of the perpendicular, and the point where the external bisector of the vertical angle meets the base produced, is equal to the square of half the sum of the sides.

E

A

ד

B

H

Let ABC be any triangle, of which AG, AH are

respectively, the perpendicular, and external bisector of the vertical angle, meeting the base in G, H: DG. DH is equal to the square of half the sum of the sides, D being the middle point of the base BC.

Produce BA, and take off AE equal to AC. Join CE, meeting AH in F. Join FG, DF.

Since AE, AC are equal, by construction, and the angles CAF, EAF are also equal; therefore (I. 4) CF, EF are equal, and AF is at right angles to CE. Hence (1) DF is parallel to BE, and equal to half of it, that is, half the sum of the sides, for BE is equal to the sum of BA, AC.

Again, the angle AHC is (I. 32) equal to the difference between the angles ACG, CAH, that is, the difference between the angles AFG, CAH (since the quadrilateral AGCF is clearly circumscribable by a circle, the angles ACG, AFG are equal (III. 21)), or, the difference between AFG, EAF, since CAH, EAF are equal, that is, between AFG, AFD (since BE, DF are parallel), which is the angle GFD. Hence (31) DG. DH = DF, or the square on half the sum of the sides.

EXAMPLES.

1. Given, of a triangle, the internal bisector of the vertical angle, the perpendicular from the vertex on the base, and the difference of the sides: construct it.

[Let (see fig. Prop. 82) AH, AG, be, respectively, the given internal bisector of the vertical angle, and the perpendicular. With these lines construct the right-angled triangle AHG. Now, since the difference of the sides is given, clearly the rectangle DG. DH is known; and the difference of the lines DG, DH, being the fixed line HG, therefore the lines DG, DH are (33) determined. Hence the middle point of the base, D, is determined. Inflect, then, DF, half the given difference of the sides, over to

AH, meeting it in F. The remainder of the solution and after proof is clear from Prop. 82.]

2. Given, of a triangle, the external bisector of the vertical angle, the perpendicular from the vertex to the base, and the sum of the sides: construct it.

[Consider Prop. 83.]

PROP. 84. To draw a common tangent to two given circles.

Let A, B be the centres of the given circles: suppose GH a common tangent to both circles, G, H being the points of contact. Join AG, BH, and draw BE parallel to HG, meeting AG in E; then BEGH is obviously a rectangle; hence EG is equal to the radius of the circle. (B), and, therefore, AE is equal to the difference between the radii of the two circles. And the angle AEB is clearly a right angle. Hence, then, the following construction :

Upon AB describe a semicircle. From A inflect over AE to the circumference, and equal to the difference of the two radii. Let AE, produced, meet the circumference of