State and prove midpoint theorem with diagram

• Triangle ABC.
• D is the midpoint of AB, so AD = DB.
• E is the midpoint of AC, so AE = EC.

Extend the line segment DE to a point F such that DE = EF. Join C to F.

• 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.

• 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).

• 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.

• 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.