Let's work on question 5A, which asks to show that the diagonals in parallelogram ABCD bisect each other.

Proof Outline:

In a parallelogram, the diagonals bisect each other. This means that the diagonals split each other into two equal parts. To prove this, we will rely on two key properties of parallelograms:

1. Opposite sides of a parallelogram are equal in length.
2. The diagonals of a parallelogram divide it into two congruent triangles.

Step-by-Step Proof:

1. Label the intersection point of the diagonals: Let the diagonals of parallelogram ABCD intersect at point O. We need to show that $AO = OC$ and $BO = OD$.

2. Consider triangles $\triangle AOB$ and $\triangle COD$:

   • Opposite sides are equal: Since ABCD is a parallelogram, we know that $AB = CD$ and $AD = BC$.
   • Angles formed by diagonals are equal: The angles formed at the intersection of the diagonals are vertical angles, so $\angle AOB = \angle COD$ and $\angle BOA = \angle DOC$.

3. Apply Side-Angle-Side (SAS) Congruence Postulate:

   • $AB = CD$
   • $\angle AOB = \angle COD$
   • $AD = BC$

   Hence, by the SAS postulate, triangles $\triangle AOB$ and $\triangle COD$ are congruent. Therefore, $AO = OC$ and $BO = OD$.

Since $AO = OC$ and $BO = OD$, the diagonals bisect each other at point O, which completes the proof.

Would you like further clarification on any steps or more details on this proof? Here are some follow-up questions related to the topic:

1. What is the definition of a parallelogram in geometry?
2. How can we use congruent triangles to prove geometric properties?
3. Why are vertical angles always congruent?
4. Can diagonals bisect each other in shapes other than parallelograms?
5. How does the property of opposite sides being equal relate to the diagonal bisecting property?

Tip: When proving properties in geometry, always start by labeling key points and identifying congruent triangles if applicable.