Part A:
Given that FG is the midsegment of isosceles triangle ABC, by the triangle midsegment theorem, we have that BF = FC and AG = GB.
Thus, BC = BF + FC = 2BF (since BF = FC)
Therefore, the measure of side BC is 2(20 mm) = 40 mm
Part B:
Similarly by the triangle midsegment theoren, GF = 1/2AC.
Given that the measure of side AC is 30 mm, therefore, the measure of side GF is 1/2(30 mm) = 15 mm
Part C:
From the given figure, EG = GF = AD.
Since, we obtained from part B that the measure of side GF is 15 mm, thus EG = GF = AD = 15 mm.
CD = CA + AD = 30 mm + 15 mm = 45 mm
Therefore, the measure of side CD is 45 mm.
Part D:
Given that EF is parallel to KM is parallel to DC, BA is a transversal such that EG = DA, this implies that line DE is parallel to line AG.
Since, line DE is parallel to line AG and KM is the midsegment of trapezoid CDEF, this implies that line KL is parallel to line DE is parallel to line AG. And ADEG forms a parallelogram with KL as the midsegment. This means that EG = KL = DA = 15 mm.
Also, LM is halfway between GF and AC, and thus the measure of line LM is the average of the measures of line GF and AC.
Thus, LM = (15 mm + 30 mm) / 2 = 45 mm / 2 = 22.5 mm
Therefore, the measure of line KM = The measure of line KL plus the measure of line LM = 15 mm + 22.5 mm = 37.5 mm
