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Answer :
Let quadrilateral ABCD be the original shape of the field.
The proposal may be implemented as follows:
Join diagonal BD and draw a line parallel to BD through point A.
Let it meet the extended side CD of
ABCD at point E.
Join BE and AD. Let them intersect each other at O.
Then, portion \(\triangle{AOB}\) can be cut from the original field so that the new shape of the field will be \(\triangle{BCE}\) as shown below.
We have to prove that the area of \(\triangle{AOB}\) (portion that was cut so as to construct Health Centre) is equal to the area of \(\triangle{DEO}\)(portion added to the field so as to make the area of the new field so formed equal to the area of the original field).
It can be observed that \(\triangle{DEB}\) and \(\triangle{DAB}\) lie on the same base BD and are between the same
parallels BD and AE.
Area (\(\triangle{DEB}\)) = Area (\(\triangle{DAB}\))
Now subtracting area (\(\triangle DOB \) ) On both side
Area (\(\triangle{DEB}\)) - Area (\(\triangle{DOB}\)) = (\(\triangle{DAB}\)) - Area (\(\triangle{DOB}\))
\(\therefore \) Area (\(\triangle{DEO}\)) = Area (\(\triangle{AOB}\))