The elements of non-Euclidean plane geometry and trigonometry, by H. S. Carslaw.

3, 4]

TWO THEOREMS

9

We shall prove that HK is perpendicular to BC and EF.
From the triangles AFA' and BFB', which are congruent,
we have AA'=BB'.
Similarly AA' =CC'. A
Therefore BB' = CC'.
Join BK and KC. B' FK \EC'
In the triangles BB'K and CC'K
we have
BB'=CC', B'K = C'K,
nL 4-1. n "nln nO4- U I n -" n cLL I ' rn nr, -II1. B 1 C.LLUL UJLLt5 &DllLJL CL U L LU. l CmL t J.UUl4t1.
Therefore the triangles are con- FIG. 7.
gruent, and BK = CK.
Again, in the triangles BHK and CHK, we have
sides equal, each to each.
Therefore the triangles are congruent, and
L BHK=LCHK=a right angle.
Also L BKH=LCKH.
But, from the triangles BB'K and CC'K, we have
/ BKB' =L CKC'.

the three

Therefore L HKB' =z HKC'=a right angle.
Thus HK is perpendicular to both BC and EF.
2. The locus of the middle points of the segments joining a set
of points ABC... on one straight line and a set A'B'C'... on
another straight line is a straight line, provided that AB = A'B',
BC = B'C', etc.

FIG. 8.

Let M, N, and P be the middle points of AA', BB', and CC'.
Join BM and produce it to B", so that BM =MB".
Join B"A' and B"B'.