Abcd is a plane quadrilateral and e is any point on ad. ef is drawn parallel to db to meet ab in f, and eg is drawn parallel to dc to meet ac in g. prove that f g is parallel to bc.

Try this:

FA/AB = AE/AD by the side-splitter theorem because EF ║ BD

∠FAE ≅ ∠BAD because an angle is congruent to itself

AE/AD = EG/DC by the side-splitter theorem because EG ║ CD

∠AEG ≅ ∠ADC because they are corresponding angles where a transversal crosses parallel lines

quadrilateral AFGE is similar to quadrilateral ABCD by SASAS similarity of quadrilaterals

From here, there are numerous ways to reach the conclusion. One of them is ...

∠AFG ≅ ∠ABC by corresponding angles of similar quadrilaterals

∴ FG ║ BC because ∠AFG and ∠ABC are corresponding angles where transversal AB crosses FG and BC

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)