AB

AC

BC

AE

EC

EF

In △ ABC, D and E are the midpoints of sides AB and AC respectively.  D and E are joined. 
Given: AD = DB and AE = EC. To Prove: DE ∥∥ BC and DE = 1/2 BC. 

Construction: Extend line segment DE to F such that DE = EF. Draw segment CF.
Proof: In △ ADE and △ CFE AE = EC  (given) 
∠AED = ∠CEF (vertically opposite angles) 
DE = EF   (construction) 
Hence △ ADE ≅ △ CFE by SAS congruence rule. 
Therefore, ∠ADE = ∠CFE  (by CPCT) and ∠DAE = ∠FCE (by CPCT) AD = CF (by CPCT).
∠ADE and ∠CFE are alternate interior angles, 
(AB and CF are 2 lines intersected by transversal DF). 
∠DAE and ∠FCE are alternate interior angles, 
(AB and CF are 2 lines intersected by transversal AC). 

Therefore, AB ∥∥ CF. So - BD ∥∥ CF. 
BD = CF (since AD = BD and it is proved above that AD = CF). 
Thus, BDFC is a parallelogram. 
By the properties of parallelogram, we have DF ∥∥ BC DF = BC DE ∥∥ BC. 
DE = 1/2BC  (DE = EF by construction). Hence proved.

A = (-6.88, -0.64)
side1length = 6 (slider from 1 to 10, step 1)
B = Point(Circle[A, side1length])
AB = Segment(A, B)
side2length = 7 (slider from 1 to 10, step 1)
side3length = 8 (slider from 1 to 10, step 1)
c = Circle(A, side2length)
d = Circle(B, side3length)
C = Intersect(c, d, 1)
AC = Segment(A, C)
k = Line(A, B)
BC = Segment(C, B)
g = Line(B, C)
D = Midpoint(AB)
E = Midpoint(AC)
DE = Segment(D, E)
α = Angle(A, E, D)
β = Angle(E, C, B)
γ = Angle(E, D, A)
δ = Angle(C, B, D)
lineFC = Line(C, AB)
h = Ray(D, E)
F = Intersect(lineFC, h)
ε = Angle(A, E, D)
ζ = Angle(C, E, F)
i = Segment(A, E)
j = Segment(E, C)
q1, b, d_1, f_1, c_1 = Polygon(B, D, F, C)
l = Segment(D, E)
m = Segment(E, F)
t1, a, e, d_2 = Polygon(E, D, A)
t2, f, c_2, e_1 = Polygon(C, E, F)
n = Segment(C, F)