Given that DE is the midsegment of ΔABC, then:
[tex]\frac{1}{2}BC\text{ = }DE[/tex]
BD = DA
AE = EC
Option A is true because:
AC = AE + EC
AC - AE = EC
Option B is true because:
AD + DB = AB
AD + AD = AB (BD = DA)
2AD = AB
AD = 1/2AB
Option C is true because:
1/2BC = DE
BC = 2DE
Given that ΔABC is scalene, then AB ≠ AC ≠ BC, in consequence, AD ≠ AE. Then option D is the correct option
