Question

E and F are the points in sides DC and AB respectively of parallelogram ABCD. If diagonal AC
and segment EF intersect at G; prove that:

Answer 1

In ∆EGC and ∆FGA

∠ECG = ∠FAG (Alternate angles as AB ∥ CD)

∠EGC = ∠FGA (Vertically opposite angles)
∆EGC ~ ∆FGA                   (By AA – similarity)

`∴("EG")/("FG")=("CG")/("AG")`

AG × EG = FG × CG