Prove that DE ∥ AC and that DE is half the length of AC .
To prove that DE || AC, we must show that the corresponding angles are equal.
Given: ABCD is a parallelogram.
Proof:
1. Since ABCD is a parallelogram, opposite sides are parallel (AB || CD and BC || AD).
2. Therefore, ∠ABC and ∠CDA are corresponding angles.
To prove that DE is half the length of AC, we can use the properties of a parallelogram and some additional lines and angles.
Proof:
1. Draw diagonal AC, bisecting ∠B and ∠C.
2. Let the point of intersection of the diagonals be E.
3. Now we have two triangles, ∆ABC and ∆ACD.
4. ∆ABC and ∆ACD are congruent triangles by SAS (side-angle-side) congruence.
a. AB = CD (opposite sides of a parallelogram)
b. BC = AD (opposite sides of a parallelogram)
c. ∠ABC = ∠CDA (corresponding angles, as shown above)
5. Therefore, AC = AC (common side).
6. By congruence, the corresponding sides in congruent triangles are equal, so AB = CD and BC = AD.
7. Since E is the midpoint of AC (by construction), AE = EC.
8. From ∆ABC and ∆ACD, we have AE + EB = EC + ED.
9. Since AE = EC, it follows that EB = ED.
10. Thus, DE is half the length of AC.
Therefore, we have proven that DE || AC and that DE is half the length of AC.