1889-90.] Mr David Maver on a Geometrical Method. 191 
CFsED(DB) = CFsEB = EBsCF(FA) = EBsCA = CAsEB(BF) = 
CAsEF = CBsEF = EFsCB(BD) = EFsCD , 
since CFsED = EFsCD, and that the generating lines ED and 
CD are equally inclined to DB the direction of motion, the angles 
EDB, CDB being supplementary, we have CF-ED = EF-CD. 
From the extremities of the base BC (fig. 5) of the triangle ABC 
are drawn any two lines BD, CE, meeting the opposite sides AC, AB, 
or those sides produced in D, E, and intersecting in F : to show that 
AB- AC-FEED = FC-FB-AE-AD. 
AB-ACEEsFD(DC) = AB-AC-FEEC = ABEEECsAC(CE) = 
ABEEECsAE = AB-FC-AEEE(EB) = ABEC-AEsFB = 
FC-AEEBsAB(BD) = FC-AE-FBsAD . 
It will be seen that the first generating line FD in the foregoing 
series of equal products is inclined to its direction of motion at the 
same angle as the last generating line AD. Hence 
AB-AC-FEED = FC-FB-AE-AD. 
ABCD (fig. 6) is a quadrilateral inscribed in a circle, the diagonals 
AC, BD intersecting at E, to show that AB 2 -CE-DE = CD 2 -AE-BE. 
The angle DCE = ABE, being in the same segment, also CDE = BAE. 
AB 2 -CEsDE(EC) = AB 2 -CEsDC = AB-CE-DCsAB(BE) = 
AB-CE-DCsAE = AB-DC- AEsCE(ED) = AB-DC-AEsCD = 
DC- AE-CDsBA(AE) = DC-AE-CDsBE = CD 2 - AEsBE , 
since the first and last generating lines DE, BE are equally 
inclined to their respective directions of motion, we have 
AB 2 -CE-DE = CD 2 -AE-BE . 
Let the opposite sides BC, AD (fig. 7) of the circumscribing quadri- 
lateral ABCD touch the circle at the points G, E. It is known 
that if GE be drawn it will pass through P, the intersection of the 
diagonals AC, BD. Show that CB-PG-AP-DP = DA-PE-PB-CP. 
CB-PG-APsDP(PA) = CB-PG-APsD A = CB-PG-DAsPA(AE) = 
CB-PG-DAsPE - CB-D A-PEsPG(GB) = CB-DA-PEsPB = 
DA-PE-PBsCB(BP) = DA-PE-PBsCP , 
