Answer: SAS
Step-by-step explanation:
Here, the given triangles AEB and ADC,
B is the mid point of the line segment AC,
⇒ AC = 2 AB,
Also, m∠A = 30°, AD = 8, ED = 4, DC = 6,
[tex]\frac{AE}{AD}=\frac{AD-ED}{8}=\frac{8-4}{8}=\frac{4}{8}=\frac{1}{2}[/tex]
[tex]\frac{AB}{AC}=\frac{AB}{2AC}=\frac{1}{2}[/tex]
⇒ [tex]\frac{AE}{AD}=\frac{AB}{AC}[/tex]
Also,
[tex]\angle EAB\cong \angle DAC[/tex]
Here, ∠EAB of triangle AEB is congruent to the corresponding ∠DAC of another triangle ADC and the sides that include this angle are proportional,
Thus, by SAS postulate of similarity,
[tex]\triangle AEB\sim \triangle ADC[/tex]
⇒ Third option is correct.