Sept., 1887.] 



ELLIOTT SOCIETY. 



179 



be limited three pairs of lines which may be called the external segments belong- 

 ing to each side ; KI giving AI and BI, belonging to side AB, KL giving BL 

 and CL, belonging to BC, KM giving CM and AM belonging to CA ; lastly 

 draw line CK. 



3. Pbop. I. First, the three perpendiculars DE, DH, and DF, are eqnal to 

 each other ; secondly, the two segments AE and AF about angle A, are equal 

 to each other, BE and BH about angle B are equal to each other, and CH and 

 CF about angle C are equal to each other. 



For, the triangles ADE and ADF have side AD common, and two angles in 

 one respectively equal to two in the other, and hence DE=:DF, and AE=AF ; 

 by like reasoning it may be shown by the triangles BDE and BDH that BH= 

 BE, and Da=DE=bF. Lastly, in the triangles CDH and CDF, we have a 

 right angle in each, the side CD common, and sides DH and DF equal ; there- 

 fore CH— CF, and CD bisects the angle C. 



Hence the perpendiculars DE, DH, DF, equal each other, the members of 

 each pair of segments about any angle are equal to each other, and the three 

 bisectrices have a common point D. 



4. Prop. II. First, the three external perpendiculars KI, KL, KM, are also 

 equal to each other ; secondly, the two segments A [ and AM about the angle 

 A are equal to each other, Bl and BL about angle B are equal to each other, 

 and CL and CM about angle C are equal to each other. 



For, the triangles AKl and AKM have side AK common, and two angles in 

 one respectively equal to two in the other, and hence KIrr KM and AI=AM ; by 

 like reasoning it may be shown in the triangles BKl and BKL that BI=BL, 

 and KL=KI=KM. Lastly, in the triangles CXL and CKM, we have a right 

 angle in each, the side CK common, and the sides KL and KM equal ; there- 

 fore CM=:CL and CK bisects the external angle at C. 



Hence the perpendiculars KI, KL, KM, equal each other, the members of 



