Point $D$ is the midpoint of median $\overline{AM}$ of triangle $ABC$. Point $E$ is the midpoint of $\overline{AB}$, and point $T$ is the intersection of $\overline{BD}$ and $\overline{ME}$. Find the area of triangle $BET$ if $[DTM] = 20$.
Let
A= (a,b)
B=(0,0)
M = (c,0)
D = [ (a +c)/2 , b/2)
E = (a/2,b/2)
DE is parallel to BM because the slope of both segments = 0
BM = c
DE = c/2
So DE = (1/2) of BM
Angle DET = Angle BMT
Angle DTE = Angle BTM
So triangle EDT is similar to triangle MBT
Since DE =(1/2) of BM.....then ET = (1/2)MT
Since triangles DET and DMT are under the same height and ET = (1/2)TM then [DET] = (1/2)[DMT ] = (1/2)(20) = 10
And DT = (1/2)BT ⇒ 2DT = BT
And since triangles DET and BET are under the same height, then [BET] = 2 [ DET] = 2 (10) = 20