Answer:
i) Therefore option A is Correct. Δ ECD [tex]\sim[/tex] Δ ACB by the SAS ( Side Angle Side) Similarity Theorem.
ii) Yes it can be proven that ED || AB after proving that Δ ECD [tex]\sim[/tex] Δ ACB
Step-by-step explanation:
i) CE = [tex]\frac{1}{2}[/tex] AC ..... given
ii) CD = [tex]\frac{1}{2}[/tex] CB .... given
iii) Therefore [tex]\frac{CE}{AC} = \frac{CD}{CB} = \frac{1}{2}[/tex]
iv) Angle ACB or ∠C is common to Δ ACB and Δ CED.
v) Therefore from the above 4 equations we can say that by
SAS theorem the two triangles are similar , that is , Δ ECD [tex]\sim[/tex] Δ ACB .
Therefore option A is Correct.
vi) Yes it can be proven that ED || AB after proving that Δ ECD [tex]\sim[/tex] Δ ACB.
Since Δ ECD [tex]\sim[/tex] Δ ACB , therefore ∠CED = ∠CAB and ∠CDE = ∠CBA.
Therefore we can say that ED is parallel to AB or that ED || AB.
