AB =CD =13 ; AD =12 ; BC =11 ; AD | | BC.
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AC =BD ==> ?
Проведем CH ⊥AD ; H∈[AD] ,
ΔCHD:
CH² =CD² -((AD -BC)/2)² ;
ΔACD:
AC² =CH²+AH² =CH²+((AD +BC)/2)² = CD² -((AD -BC)/2)² +((AD +BC)/2)² = CD² +AD*BC ;
AC² = CD² +AD*BC ;
AC² = 13² +12*11 ;
AC=√301