In trapezoid $ABCD$, $\overline{AB}$ is parallel to $\overline{CD}$, $AB = 7$ units, and $CD = 10$ units. Segment $EF$ is drawn parallel to $\overline{AB}$ with $E$ lying on $\overline{AD}$ and $F$ lying on $\overline{BC}$. If $BF:FC = 3:4$, what is $EF$? Express your answer as a common fraction.

Cut and paste does not work here, as you can see, but I think I followed your coding.

extend both CB and DA to meet at P
let FB = 3x, let CF = 4x, let BP= a, let FE = k
(we want k)

you now have 3 similar triangles
PBA, PFE, and PCD

a : a+3x : a+7x = 7 : k : 10

a/7 = (a+7x)/10
10a = 7a + 49x
3a = 49x
a = 49x/3

a/7 = (a+3x)/k
ak = 7a + 21x
k = (7a +21x)/a
= (7(49x/3 + 21x)/(49x/3)
= ( 343x/3 + 21x)/(49x/3)
= (406x/3) / (49x/3)
= 406/49
= 58/7

so EF = 58/7 or appr 8.29

check my arithmetic

Reiny's answer is correct, and I did check her/his computation.

Stop cheating!

www.jiskha.com/questions/1798144/point-g-is-the-midpoint-of-the-median-xm-of-xyz-point-h-is-the-midpoint-of-xy-and-point

Stop Cheating. Really?

wow. People just need some help.

Really, ur aops???

Thats just dumb

yea i highly doubt that. Nice try tho. And in the highly unlikely chance u are AOPS, well, I went to the link YOU ARE NOT AOPS posted, and that's just pathetic

you are not AoPS, however, it is not right to cheat. anyways stop being an imposter.

pls don't cheat, but don't pose as AoPS we know ur faking fs

yep its 58/7

So what if they're cheating? It'll come back to hurt them later in life.