angles E and B are equal; also the angles F and C are equal; therefore the remaining angle D is equal to BAC (I. 32), and therefore the triangle BAC inscribed in the given circle is equiangular to the given triangle EDF.

PROPOSITION III.-PROBLEM.

Προτασις γ ́.

Περὶ τὸν δοθέντα κύκλον τῷ δοθέντι τριγώνῳ

ἰσογώνιαν τρίγωνον περιγράψαι.

About a given circle ABC to circumscribe a triangle equiangular to a given triangle EDF.

Construction.-Produce any side DF of the given

triangle both ways to G and H; from the centre K of the given circle draw any right line KA; with this line at the point K make the angle BKA equal to the angle EDG, and at the other

side of KA make the angle AKC equal to EFH; and draw the lines LM, LN, and MN tangents to the circle in the points B, A, and C.

Proof. Because the four angles of the quadrilateral figure LBKA taken together are equal to four right angles (I. 32), and the angles KBL and KAL are right angles (III. 18), the remaining angles AKB and ALB are together equal to two right angles; but the angles EDG and EDF are together equal to two right angles (I. 13); therefore the angles AKB and ALB are together equal to EDG and EDF; but AKB and EDG are equal by construction, and therefore ALB and EDF are equal. In the same manner it can be demonstrated that the angles ANC and EFD are equal: therefore the remain

ing angle M is equal to the angle E (I. 32), and therefore the triangle LMN circumscribed about the given circle is equiangular to the given triangle.

PROPOSITION IV.-PROBLEM.

Πρότασις δ'.—Εἰς τὸ δοθέν τρίγωνον κύκλον ἐγγραψαι.
To inscribe a circle in a given triangle BAC.

Construction.—Bisect any two angles B and C by the right lines BD and CD, and from their point of concourse D draw DF perpendicular to any side BC; the circle described from the centre D with the radius DF is inscribed in the given triangle. Draw DE and DG perpendicular to BA and AC.

B

F

Proof. In the triangles DEB, DFB, the angles DEB and DBE are equal to the angles DFB and DBF by construction, and the side DB is common to both, therefore the sides DE and DF are equal (I. 26): in the same manner it can be demonstrated that the lines DG and DF are equal; therefore the three lines DE, DF, and DG, are equal; and therefore the circle described from the centre D with the radius DF passes through the points E and G; and because the angles at F, E, and G, are right, the lines BC, BA, and AC, are tangents to the circle (III. 16); therefore the circle FEG is inscribed in the given triangle.

Annotations.

Corollary 1.-If the sides be given in numbers, the radius of the Inscribed circle may be found by dividing the area of the triangle by the semi-perimeter; because the whole triangle is composed of three triangles, whose bases are the sides, and common altitude the radius of inscribed circle.

Corollary 2.-The bisectors of the angles of a triangle intersect in the same point; which is the centre of the inscribed circle.

Corollary 3.-The bisectors of any internal angle, and of the remaining two external angles, intersect in the same point, which is the centre of the circle touching the side opposite the given angle, and the productions of the two sides containing it.

The circles thus described are

said to be exscribed to the triangle; thus if BD and CD be drawn bisecting the external base angles of the triangle ABC, to meet in D; it can be proved, as in the Proposition, that DA bisects the vertical angle, and that perpendiculars DF, DG, and DE, let fall upon the sides are all equal. The circle EFG is said to be exscribed to the triangle ABC at the side BC.

Corollary 4.-The radius of the exscribed circle may be found in numbers, by dividing the area of the triangle by half

the difference between the sum of the two sides and the base; because the whole triangle is the difference between the sum of the two triangles ACD and ABD, whose bases are the sides, and common altitude the radius of exscribed circle, and the triangle BCD, whose base is the third side, and altitude the same radius.

Corollary 5.-From the preceding Corollary, and Corollary 4 (II. 13), we can express readily the radii of the three exscribed circles. Let r', r", "", denote the radii of the circles exscribed to the sides opposite A, B, and C, respectively; we have

If r denote the radius of inscribed circle, we have also, by Corol. 1,

Corollary 6.-By multiplying together the values of the four radii found in the preceding Corollary, we obtain the following remarkable expression for the area of a triangle,

Area =Vrr'r"r".

EXERCISE.

Find the radii of inscribed and exscribed circles in the triangle, whose sides are, a 100, b = 86, c=72; hence,

If we multiply the first four of these quantities together, and extract the square root of the product, we find, as in the 1st Example of Corol. 4 (II. 13).

Area 3028.06.

If we multiply together the last four quantities, we find.

Area 3028.04.

=

Corollary 7.-In a right-angled triangle, the diameter of the inscribed circle is equal to the difference between the sum of the sides and the hypotenuse; and the diameter of the circle exscribed to the hypotenuse is equal to the perimeter of the triangle.

PROPOSITION V.-PROBLEM.

Προτασις εʹ.-Περὶ τὸ δοθὲν τρίγωνον κύκλον περιγράψαι.
To circumscribe a circle about a given triangle BAC.
Construction.-Bisect any two sides

BA and AC of the given triangle; and
through the points of bisection D and
E draw DF and EF perpendicular to
AB and AC; and from their point of
concourse F draw FA, FB, and FC; B
the circle described from the centre F

with the radius FA is circumscribed about the given triangle.

Proof. In the triangles FDA, FDB, the sides DA and DB are equal by construction, FD is common to

both, and the angles at D are right; therefore the sides FA and FB are equal (I. 4). In the same manner it can be demonstrated that the lines FA and FC are equal, therefore the three lines FA, FB, and FC are equal; and therefore the circle described from the centre F with the radius FA passes through B and C, and therefore is circumscribed about the given triangle BAC.

ΠΟΡΙΣΜΑ.

COROLLARY.-If the centre F fall within the triangle, it is evident that all the angles are acute, for each of them is in a segment greater than a semicircle (III. 31). If the centre F be in any side of the triangle, the angle opposite to that side is right, because it is an angle in a semicircle. And if the centre fall without the triangle, the angle opposite to the side. which is nearest to the centre is obtuse, because it is an angle in a segment less than a semicircle.

Annotations.

Corollary 1.- -The three straight lines which bisect the sides of a triangle at right angles to them meet in the same point, which is the centre of the circumscribing circle.

Corollary 2.-If the three points of bisection of the sides of a triangle be joined, the perpendiculars to the sides at the points of bisection are also perpendiculars to the sides of the triangle so formed; therefore, if in any triangle perpendiculars be let fall from each angle on the opposite side, these perpendiculars intersect each other in the same point.