Question

1. identify all pairs of parallel segments.

** this is a 2-pag
a)

b)

c)

midpoints of the side

Explanation:

Step1: Recall the mid - segment theorem

The mid - segment of a triangle is parallel to the third side. In \(\triangle ACE\), \(B\) is the mid - point of \(AC\) (since \(AB = BC\)), \(F\) is the mid - point of \(AE\) (since \(AF=FE\)). So, \(BF\) is a mid - segment.

Step2: Apply the mid - segment theorem for \(\triangle ACE\)

By the mid - segment theorem, \(BF\parallel CE\).

Step3: Consider \(\triangle ACE\) again for another mid - segment

In \(\triangle ACE\), \(D\) is the mid - point of \(CE\) (since \(CD = DE\)), \(F\) is the mid - point of \(AE\) (since \(AF = FE\)). So, \(FD\) is a mid - segment.

Step4: Apply the mid - segment theorem for the second mid - segment in \(\triangle ACE\)

By the mid - segment theorem, \(FD\parallel AC\).

Step5: Consider \(\triangle CEA\) for the third pair

In \(\triangle CEA\), \(B\) is the mid - point of \(AC\) (since \(AB = BC\)), \(D\) is the mid - point of \(CE\) (since \(CD=DE\)). So, \(BD\) is a mid - segment.

Step6: Apply the mid - segment theorem for the third mid - segment in \(\triangle CEA\)

By the mid - segment theorem, \(BD\parallel AE\).

Answer:

a) \(BF\parallel CE\)
b) \(FD\parallel AC\)
c) \(BD\parallel AE\)