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.
If you extend both BC and AD to meet at P you get 3 similar triangles PAB, PEF, and PCD. Let's call BF = 3x and FC = 4x. Therefore the following proportions can be expressed:
It would be 4x+76 = 180 in that order in the boxes. You would add 1 to 75 which got 76. Then you would subtract 76 from 180 and get 104 and then divide that by 4 and you would get 26. So x is 26
The like terms are 12 and 5, so you have to subtract 5 from both sides. 12 < x + 5 -5. -5 7 < x We can just switch it around so x is in the left side. x > 7
