AC, DP, AL, BC, DG, BD, EP, AG, DF, AE, AH, GL, DK, on one side, cancel the same quantities on the other side, we have BI × FH× GK = IF BG HK; whence, by Theorem IX., the × × three points G, H, and I are in the same straight line, the triangle whose sides are cut being KBF.

NOTE.—The interesting property involved in this theorem was contained in a problem proposed for demonstration about the year 1830, in the last number of the Mathematical Diary, a monthly periodical, commenced by Prof. Robert Adrain in 1825, and afterwards conducted by James Ryan, of New York; but, the work being discontinued, no demonstration was published.

THEOREM XI.*-If in each of the three sides of a plane triangle a point be taken at pleasure, and circles be described through each angular point of the triangle, and the points taken in the two sides forming that angle, these three circles will intersect one another at a common point. Required, the demonstration.

Demonstration.-Join DP, EP, and FP. Then, since ADPF is a quadrilateral inscribed in a circle, its opposite angles A and P are equal to two right angles (III. 18, cor. 4†). For the same reason, the opposite angles C and P of the quadrilateral FCEP are equal to two right angles. Whence these four angles A, C, DPF, and FPE are equal to four right angles. But the three angles DPF, FPE, and DPE are equal to four right angles. Whence the angles A + C + DPF + FPE DPF FPE + DPE. Wherefore DPE A+ C. To each add the angle B, and we have DPE + B = A + C + B = two right angles; consequently the points D, P, E, and B, which designate the quadrilateral DPEB, are in the

*In seeking a mode of demonstration for Theorem XIV., following, which was sent to me some years ago by my valued cousin and mathematical correspondent, Benjamin Shoemaker, now of Germantown District, Philadelphia, who was my student in 1826, I discovered the interesting properties in Theorems XI., XII., and XIII. The same properties may, however, have been previously discovered by some one with whose writings I have not had the pleasure to meet.

† III. 22.

circumference of a circle, and hence the circumference of a circle which passes through the three points D, B, and E must pass through the point P, the intersection of the circles through the points A, D, F, and F, C, E. Q. E. D.

Corollary. The angle DPF the sum of the angles B and C, the angle FPE=the sum of the angles A and B, and the angle DPE the sum of the angles A and C.

THEOREM XII.-If in each of the three sides of a plane triangle any point be taken at pleasure, and a circle be described through each angular point of the triangle and the two points taken in the sides that form that angle, the triangle formed by joining the centres of these three circles will be equiangular and similar to the first triangle.

Demonstration.-The arc LP (III. 11*) is half the arc FLP, and the arc PS is half the arc PSD; hence the whole arc LPS, which measures the angle H, is half the whole are FPD. But (III. 18†) the angle A is measured by half the same are FPD; hence the angle H of the triangle HIG is equal to the angle A of the triangle ABC.

In like manner the arc RP is half the arc DRP, and the arc PO is half the arc POE; hence the arc RPO, which measures the angle G, is half the arc DPE. But the angle B is measured by half the arc DPE; therefore the angle G of the triangle HIG is equal to the angle B of the triangle ABC.

So, also, arc KP FKP, and arc PQPQE; hence are KPQ, which measures the angle I, is equal to half FPE. But FPE measures the angle C. Therefore the angle I of the triangle HIG is equal to the angle C of the triangle ABC. Consequently, the two triangles HIG and ABC are equiangular and similar. Q. E. D.

THEOREM XIII-If in each of the three sides of any plane triangle any point be taken at pleasure, all triangles whose sides pass through these points and terminate in the circular arcs passing through the angular points of the triangle and the points taken in the sides which form that angle, will be equiangular and similar.

Demonstration.-Through the point D draw any line KDL, and through the points F and E draw the lines KF and LE, and let them be produced to meet in a point, as M. Then, since in the triangles ABC and KLM (III. 18, cor. 1*) the angle Kangle A, they being in the same segment, and the angle L the angle B for the same reason, the angle M must be equal to the angle C, and consequently the point M is in the circular are FCME, and the triangle KML is equiangular and similar to ABC. Q. E. D.

Corollary.—Join the centres H, I, G of the circular arcs forming the triangle HIG; then, as the triangle HIG, by Theorem XII., is similar to ABC, all the triangles formed, as KLM, will be similar to HIG and ABC, and hence similar to one another.

Scholium 1.- Of all the sides drawn through one of the points, as F, and terminating in the circular arcs FCE and DAF that intersect at that point, the longest or maximum line will be that (KFM) drawn parallel to the line HI, which joins the centres of those arcs. For let any other line, as AFC, pass through F, and on it let fall the perpendiculars HP and IQ. Also, on KM let fall the perpendiculars HN and IO, and draw HR parallel to AC. Then, in the right-angled triangle HRI, HR is less than the hypothenuse HI. Now (III. 6†) AF 2 PF, and FC 2 FQ; hence AC-2 PQ= 2 HR. Also, KF=2 NF, and FM- 2 FO; whence KM-2 NO 2 HI. And, since HI is greater than HR, we have KM (— 2 HI) greater than AC (2 HR). Q. E. D.

F

R

C

M

Scholium 2.-Join KD and ME, and produce them to meet in L. Then, by the theorem, the point L will be in the circular arc DBE, and the triangle KLM will be similar to ABC. But KM is longer than any other side, as AC, drawn through F Hence the area of the triangle KLM is greater than ABC, or than other triangle whose sides pass any through the points D, E, F, and terminate in the circular arcs DAF, FCE, and EBD, and it is therefore the maximum triangle which can be drawn similar to ABC, and having

N

H

K

A

D

G

E

L

B

its sides to pass through the points D, E, and F.

Scholium 3.—If the lines KL, LM, and MK be drawn through the points D, E, and F, respectively parallel to the sides HG, GI, and II of the triangle GIH, the angular points K, L, and M will be in the circular arcs DAF, FCE, and EBD.

For (IV. 21*) the angles L, M, and K in the triangle LMK are respectively equal to the angles G, I, and H in the triangle GIH. But, by Theorem XII., the angles G, I, and H are respectively equal to the angles B, C, and A of the triangle ABC. Hence the angle I must equal the angle B, the angle Mangle C, and angle K— angle A, and the point K must be in the circular arc DAF, M in the arc FCE, and L in the arc EBD.

Scholium 4.-Hence the maximum triangle KLM has its sides parallel, respectively, to the sides of the triangle HIG, formed by joining the centres of the circular arcs DAF, FCE, and EBD.

*VI. 5 and 4.

THEOREM XIV. (Proposed by Benjamin Shoemaker.)-If on the three sides of any plane triangle equilateral triangles be described, the triangle formed by joining the centres of these equilateral triangles will be an equilateral triangle.

Demonstration.-On the three sides

M

E

F

G

C

P

K

H

'D

I

B

of the triangle DEF describe the equilateral triangles DBF, FAE, and ECD. Bisect the sides BD and DF in the points O and P, and join the points FO and BP by lines intersecting at H; then will I be the centre of gravity of the triangle DBF, and hence its centre. But FO and BP are perpendiculars from the middle of the lines BD and DF respectively, and hence their intersection H is the centre of a circular are passing through the points. D, B, and F. With the centre H and radius HB or HF describe the arc DBF. In like manner describe the arcs FAE and ECD, whose centres I and G will be the centres of the triangles FAE and ECD respectively. Join the points H, I, and G; it remains to be shown that HIG is an equilateral triangle.

Through the point D draw any line, as KDL, and draw KFM and LEM, and the point M, where they meet, will, by Theorem XIII., be in the arc FAE, and the triangle KLM will be equiangular to HIG (cor. to Theorem XIII.). But angle K (III. 18, cor. 1*) angle B = an angle of an equilateral triangle. For a similar reason the angles M and L, which are equal to the angles A and C, are each an angle of an equilateral triangle. Hence KLM is an equilateral triangle, and, consequently, the triangle HIG, which is similar to KLM (Theorem XII.), is an equilateral triangle also. Q. E. D.

Scholium.—This theorem resolves into Theorem XII., wherein the given triangle KLM, corresponding to ABC in Theorem XII., is equilateral.

K*

* III. 21.