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How to prove the statement " If a parallelogram has a circumscribed circle, then it is a rectangle " and its converse
" If a parallelogram is a rectangle, then it has a circumscribed circle ".
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1. " If a parallelogram has a circumscribed circle, then it is a rectangle ".
Let a parallelogram ABCD has a circumscribed circle.
Then, since in a parallelogram the opposite sides are congruent, these arcs are congruent in pairs:
arc(AB) ~ arc(CD) and arc(BC) ~ arc(DA) ("~" means "congruent arcs").
Then the arcs arc(AB) + arc(BC) and arc(CD) + arc(DA) are congruent:
arc(AB) + arc(BC) ~ arc(cd) + arc(DA).
From the other side, their sum
(arc(AB) + arc(BC)) + (arc(CD) + arc(DA))
is the entire circle, i.e 360.
It implies that
arc(AB) + arc(BC) = 180 and
arc(CD) + arc(DA) = 180.
Then each of the two angles ABC and CDA is the right angle, since each of the angles leans the 180 arc.
Thus we proved that the given parallelogram has at least one (actually, two) right angles.
Hence, the parallelogram is a rectangle.
Thus the statement #1 is proved.
2. " If a parallelogram is a rectangle, then it has a circumscribed circle ".
This statement is easier to prove.
a) Since in any rectangle the diagonals are congruent, and
since in any parallelogram the diagonals bisect each other,
it implies that the intersection point of diagonals of the rectangle is equidistant from its verices.
Thus the intersection of diagonals is the center of the circle circumscribed about the rectangle.
Proved and solved.
You have this free of charge online textbook on Geometry
GEOMETRY - YOUR ONLINE TEXTBOOK
in this site.
The relevant sections of this textbook are
Properties of parallelograms and
Properties of circles, inscribed angles, chords, secants and tangents
The lessons in these sections cover all properties I used in the proof.
H a p p y l e a r n i n g !