State and prove midpoint theorem with diagram

Here's the Midpoint Theorem, its diagram description, and proof: Midpoint Theorem The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is half the length of the third side. Diagram Description Imagine a triangle labeled ABC. Let D be the midpoint of side AB. Let E be the midpoint of side AC. Draw a line segment connecting D and E. This line segment is DE. Proof Let's prove the theorem using geometric construction and congruence. 1. Given: Triangle ABC. D is the midpoint of AB, so AD = DB. E is the midpoint of AC, so AE = EC. 2. Construction: Extend the line segment DE to a point F such that DE = EF. Join C to F. 3. Step 1: Prove ΔADE ≅ ΔCFE Consider triangles ΔADE and ΔCFE: AE = EC (Given, E is the midpoint of AC) ∠AED = ∠CEF (These are vertically opposite angles*) DE = EF (By construction) Therefore, ΔADE ≅ ΔCFE by the SAS (Side-Angle-Side) congruence criterion*. 4. Step 2: Deduce properties from congruence Since ΔADE ≅ ΔCFE, their corresponding parts are equal (CPCTC*). AD = CF (Corresponding sides) ∠DAE = ∠FCE (Corresponding angles) The equality of alternate interior angles (∠DAE and ∠FCE) implies that AB is parallel* to FC (AB || FC). 5. Step 3: Show BCFD is a parallelogram We know AD = DB (Given, D is the midpoint of AB). From Step 2, we found AD = CF. Therefore, DB = CF. We also established that AB || FC, which means DB || FC. Since one pair of opposite sides (DB and CF) are equal and parallel, the quadrilateral BCFD is a parallelogram*. 6. Step 4: Conclude the theorem Because BCFD is a parallelogram, its opposite sides are parallel and equal. Therefore, DF || BC and DF = BC. We know that DF = DE + EF. By construction, DE = EF, so DF = DE + DE = 2DE. Substituting this into DF = BC, we get 2DE = BC. This implies DE = (1/2)BC*. Also, since DF || BC, it means DE || BC*. Thus, the line segment joining the midpoints D and E is parallel to the third side BC and is half its length.