Triangulating Text

Python, 81 bytes

def f(s,p=''):
 i=-int(len(2*s)**.5)
 if s:f(s[:i],p+' ');print p+' '.join(s[i:])

A recursive function. Goes from the end of s, chopping off and printing characters. The number of characters to take is computed from the length of s. The function is set up to prints in reverse order of the recursive calls, which terminate when s is empty and then resolve back up the line. Each layer, the prefix p has an extra space added.

In Python 3, the if can be done via short circuiting, though this doesn't seem to save chars:

def f(s,p=''):i=-int(len(2*s)**.5);s and[f(s[:i],p+' '),print(p+' '.join(s[i:]))]

An equally-long alternative with inequality chaining:

def f(s,p=''):i=-int(len(2*s)**.5);''<s!=f(s[:i],p+' ')!=print(p+' '.join(s[i:]))

Both print and f return None, which is hard to use.

Pyth, 22 bytes

jua+L\ GjdHfTczsM._UzY

Try it online: Demonstration or Test Suite

Explanation:

jua+L\ GjdHfTczsM._UzY   implicit: z = input string
                   Uz    create the list [0, 1, ..., len(z)-1]
                 ._      all prefixes of this list: [[0], [0,1], [0,1,2], ...]
               sM        sum up each sublist: [0, 1, 3, 6, 10, ...]
             cz          split z at these indices
           fT            remove all the unnecessary empty strings
                         this gives us the list of strings of the triangle
 u                   Y   reduce this list, with the initial value G = []
   +L\ G                    prepend a space to each string in G
        jdH                 join the current string with spaces
  a                         and append it to G
j                        print each string on a separate line

Candy, 67 59 57 bytes

&iZ1-=yZ1+Z*2/>{0g}0=z@1i&{|.}bYR(" ";=)ZR(=a&{;}" ";)"\n";Y1-=ya1j

&1-8*1+r1-2/=y@1i&{|.}bYR(" ";=)ZR(=a&{;}" ";)"\n";Y1-=ya1j

&8*7-r1-2/=y@1i&{|.}bYR(" ";=)ZR(=a&{;}" ";)"\n";Y1-=ya1j

or:

          &
         8 *
        7 - r
       1 - 2 /
      = y @ 1 i
     & { | . } b
    Y R ( "   " ;
   = ) Z R ( = a &
  { ; } "   " ; ) "
 \ n " ; Y 1 - = y a
1 j

long form:

stackSz
digit8    # Y = (sqrt((numCh - 1) * 8 + 1) - 1) / 2   using pythagorean
mult      # Y = (sqrt(numCh * 8 - 7) - 1) / 2  equivalent but shorter
digit7
sub
root
digit1
sub
digit2
div
popA
YGetsA
label digit1
incrZ
stackSz   # bail if we're out of letters
if
  else
  retSub
endif
stack2
pushY     # print the leading spaces (" " x Y)
range1
while
  " " printChr
  popA
endwhile
pushZ
range1      # output this row of characters (Z of them)
while
  popA
  stack1
  stackSz
  if
    printChr    # bail on unbalanced tree
  endif
  " " printChr
endwhile
"\n" printChr
pushY
digit1
sub
popA
YGetsA
stack1
digit1 jumpSub   # loop using recursion