Given: \overline{AD} \cong \overline{AE} AD ≅ AE and \overline{AB} \cong \overline{AC}. AB ≅ AC . Prove: \triangle ABE \cong \triangle ACD△ABE≅△ACD.

May consider a group  g=({0,1},+)g=({0,1},+)where  ++  will work as the xor operator like we work in boolean algebra. note here, that ‘00’ and ‘11’ are not the usual 0 and 1 that we work with everyday. the elements are  called  ‘00’ and ‘11’ and  may not mean  zero (nothing) and one (single). ‘00’ is the identity for the given binary operator +, meaning that any element operated with ‘00’ gets us that element itself.  for a set  ss  in  g′g′  having more elements than just  {0,1}{0,1}  the values for  a+b∈sa+b∈s, such that  a∈s,b∈sa∈s,b∈s, have to be first defined, only then can you determine the value of  1+11+1. the answer could possibly have been  22  if oanswer is25,000

Answerbecause of improved transportation, communication, and other advantages;

Answerwill have thinner; have thicker oceanic sediments;

Ithink is a i believe