In triangle ABC, AB = AC = 25 and BC = 20. Points D, E and F are on sides AB, BC and AC respectively, such that DE and EF are parallel to AC and AB respectively. What is the perimeter of parallelogram ADEF?
+171 i mean $ \overset{-}{AD}=\overset{-}{FE} $
knowing all that we can say that $\overset{-}{DB}$ and $ \overset{-}{DE} $ form an isosceles triangle so yes logically we get $\overset{-}{AB}=\overset{-}{DE}+\overset{-}{FE}=25 $
for the other side we can say that $ \overset{-}{AF} $ and $\overset{-}{DE}$ have the same measurement -- so does $\overset{-}{FE}$ and $\overset{-}{FC}$ -- thus we can say that $\overset{-}{AC} = \overset{-}{DE} + \overset{-}{FE} = 25$
so yes, since we have the two sides just add them together
although may i know why i cannot edit the original answer?
+171
look at the image -- since $ \overset{-}{AD} = \overset{-}{AC} $ then we can say that $ \overset{-}{AB}=\overset{-}{DE}+\overset{-}{FE}=25 $
its the exact same from the other point of view, so its P would be $ 25+25 $, which is equal to 50
+171 i mean $ \overset{-}{AD}=\overset{-}{FE} $
knowing all that we can say that $\overset{-}{DB}$ and $ \overset{-}{DE} $ form an isosceles triangle so yes logically we get $\overset{-}{AB}=\overset{-}{DE}+\overset{-}{FE}=25 $
for the other side we can say that $ \overset{-}{AF} $ and $\overset{-}{DE}$ have the same measurement -- so does $\overset{-}{FE}$ and $\overset{-}{FC}$ -- thus we can say that $\overset{-}{AC} = \overset{-}{DE} + \overset{-}{FE} = 25$
so yes, since we have the two sides just add them together
although may i know why i cannot edit the original answer?
