114 
Mr, W HE WELL on calculatmg 
S3. Prop. Having given the symbol of a plane derived 
from the 'tetrahedron, to find the manner in which it cuts the 
cKtahedron, Fig. i 3 - 
Let PQR be any plane at the angle A ; and let PQ meet 
DKandDE in S andT /. DS= 
=DP . ^ =DP . 
AP ha q 
And drawing QL parallel to DE, DF = 
AlsoQL= AO and PL = AP — AL = AP — AQ . ^ = 
— kc 
.•.DT = DP.^ = DP. 
V 
9-P 
c 
a 
In the same way we find the portions cut off from DH and 
DF : and hence it appears that a plane ( ^ 
from the four edges, which meet at the vertex D of the pyra- 
mid, lines which, parallel to the edges in the directions Ay, 
A z, xy, X Zy are as 
III I 
q ’ r ’ q — r — p 
In whatever manner the 'plane DEF is cut by the plane 
PQR. the plane DHK will be similarly cut by the co-existent 
plane at x. 
34. Hence, knowing the law by which~a secondary face is 
derived from the octahedron, we can find its symbol. 
The primary form is a ‘ square octahedron ; to find the 
symbol of the face (Ex. Zircon unihinaire, Hauy). 
This plane is drawn cutting off the angle E, in such a man- 
ner that the portions cut from E F, E G are double of those 
from EK, ED respectively ; and the section on the face EFG 
parallel to FG or to Ay. 
Since the part cut from EG, parallel to A z, is double of that 
