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?

Guest Jun 24, 2021

#2**+2 **

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?

UsernameTooShort Jun 24, 2021

#1**+2 **

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

UsernameTooShort Jun 24, 2021

#2**+2 **

Best Answer

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?

UsernameTooShort
Jun 24, 2021