Answer:
Option B is correct.
Step-by-step explanation:
We have given a triangle ABC and EDC please look at the figure
We can see that AE and BD are transversal therefore, â EAB=â AED being alternate interior angles
And â ACB=â DCE are vertically opposite angles hence, equal
So, by AA similarity postulate the above to triangles are similar
ÎABC [tex]\sim[/tex] ÎEDC
Therefore, Option B is correct that is Triangle ABC is similar to triangle EDC , because mâ 3 = mâ 4 and mâ 1 = mâ 5
NOTE: mâ 3 = mâ 4 corresponds to mâ ACB=mâ DCE
And mâ 1 = mâ 5 corresponds to mâ EAB=mâ AED