Question

Please help if you can. Will give brainliest.

Answer 1

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,

$\frac{AE}{AD}=\frac{AD-ED}{8}=\frac{8-4}{8}=\frac{4}{8}=\frac{1}{2}$

$\frac{AB}{AC}=\frac{AB}{2AC}=\frac{1}{2}$

⇒  $\frac{AE}{AD}=\frac{AB}{AC}$

Also,

$\angle EAB\cong \angle DAC$

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,

$\triangle AEB\sim \triangle ADC$

⇒ Third option is correct.