286 Mr. T. K, Abbott on a Geometrical Problem. 



The present analysis also differs widely from one by Mr. R. 

 Pattison* of an incrustation from the volcanic springs of New 

 Zealand (but of which the particular locality is not given), as 

 his specimen yielded — 



Silica 77-35 



Alumina 9-70 



Peroxide of iron . . 3*72 



Lime 1'74 



Water 7*66 



10017 



The occurrence of chlorine in deposits of this character has 

 not, as far as I am aware, been before noticed, though potash 

 and soda in small quantities have been detected, existing, it 

 seemed probable, in combination with the silica. The existence 

 of chloride of sodium, therefore, in this incrustation appears to 

 be a point of some interest, though its bearing upon the chemical 

 geology of these volcanic springs could not be considered without 

 more distinct information as to the nature of the springs them- 

 selves, and the circumstances under which the incrustation was 

 formed, than I have been able to obtain. 



XLIV. On a Geometrical Problem noticed by Mr. Sylvester. 

 By Thomas K. Abbott, Esq.f 



IN answer to Mr. Sylvester's invitation, I beg to offer a direct 

 demonstration of the theorem, that if from the middle a of 

 a circular arc be chords ad, ae be drawn whose remote segments 

 md, ne are equal, the whole chords are so. The rectangle en . na 

 = cn.nb. Add an^=no--\-oa^ [o being middle of be); then 

 ea . na-=.oc^ -\- ao^—ac^ . Similarly for the other chord, .*. ea.na 

 •=.da .ma. Bisect me? in / and we in ^, add to both rectangles 

 mp=ng^'y then we have ap=^ag'^, .*. af^ag, ,\af->tfd=zag-\-gc, 

 Q. E. D. 



Now to show directly that if the bisectors of the base angles 

 of a triangle be equal the sides are equal. Let the reader form 

 the figure thus : bd bisecting < abc, a/" bisecting < bac ; let them 

 meet in o, and let the external bisector of c meet the former in e 

 and the latter in g. Then we have the harmonic mean between 

 ao, ag = that between bo, be, .*. producing be to c', so that 

 ed = bo and ag to g', so that gg^ — ao ; and reciprocating the figure 

 from the origin o, the points b, a, ef, g' give a parallelogram cir- 

 cumscribing a circle ; and if through p, q the feet of perpendi- 



* Phil. Mag. vol. xxv. p. 496. 

 t Communicated by the Author. 



