Fg is the midsegment of isosceles triangle abc, and km is the midsegment of trapezoid cdef. ef || km || dc. find the following measures. show your work and describe any properties or theorems you use.

Well, you didn't include the shape, but I believe I already know it, I've actually answered this problem before:

Point FG is located at half distance down the side of the triangle. BC would be twice as much as that of BF, and BF is equal to 20 according to the problem. 20*2=40, so BC=40.

GF is at the bottom, and is half the size of ABC, Gf is thus half the size of AC, making GF=15.

GF and GE are the same lengths, so GE is 15 as well. DA is the same as GE, so this means DA is also 15. If AC is 30, and DA is 15, then 30+15=45. Making CD=45.

KM is at the midpoint of DEFC, making LM be halfway between GF and AC. GF was 15 and AC was 30, you must find the average. LM=22.5

KL is parralel to DA and KD is parralel to LA, DKLA must be a parralelogram. KL is the same as DA, which is the same as GF and GE. 

We know LM is 22.5, and if KL is also 15, then KM=22.5+15, 37.5

BC=40

GF=15

CD=45

KM=37.5

Step-by-step explanation:

answer: 75°

step-by-step explanation:

∠esk is the supplement of ∠edk, so is 100°. ∠esd = 1/2·arc ed, so is 25°.

since ∠esk = ∠esd + ∠dsk

you have 100° = 25° + ∠dsk . .

fill in known values 75° = ∠dsk . . . . .

subtract 25°

step-by-step explanation:

what am i looking for finish the question.

step-by-step explanation: