Question

E and F are the midpoints of the sides AB and CD respectively of the parallelogram ABCD. In each of the following, if the statement is true, give reason 1. DF=EB 2. AF=EC​

Answer 1

Answer:

I don't have a feature to send pics so sorry for that

Step-by-step explanation:

here we have ABCD as parallelogram and E&F are mid points of AB &CD.

Since we know by the property of parallelogram

AB=CD and AD=BC

Now by mid points we got

AE=EB and CF=FD

now again by property of parallelogram we said that

AE=DF=EB=FC

=>{DF=EB}

Now by above 2nd condition

we got two new lengths as AF & EC and also make two triangle as ADF and BEC

Now to prove AF=EC, first we've make those triangles congrent,

1st: AD=BC (AD from ADF and BC from BCE)

2nd: /_ADF=/_CBE (/_ADF from ADF and /_CBE from BCE)

3rd: /_AFD=/_CEB (". ". ". ". ". ". ". " )

So by AAS property

triangleADF≈ triangle CEB

so by this we said that {AF=EC}

here the 3rd point is taken from figure of parallelogram bit I don't have features so I can't send the pics of solution