## Area of a quadrilateral II

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• Once I ve known the right formula (Blaschke s formula) from Sergei Markelov, my proof has become much easier. I send it here hoping it could be of some
Message 1 of 1 , Sep 30, 2002
Once I've known the right formula (Blaschke's formula) from Sergei
Markelov, my proof has become much easier.
I send it here hoping it could be of some interest for someone.
This is the proof in the general case, non only for cyclic

perpendiculars in the vertexes to their own bisectors (SP is
perpendicular in A to the A-bisector, PQ is perpendicular in B to the
B-bisector,).
With some angle considerations SPQR results to be cyclic.
All quadrilaterals inscribed in SPQR having their sides parallel to
those of ABCD have the same perimeter of ABCD (this can be seen with
trigonometric considerations or in I. M. Yaglom's Geometric
Transformations II, solution to problem n. 76).
It is clear that the distance between the parallel sides of any two
quadrilaterals A'B'C'D' and A''B''C''D'' so constructed is the same:
I mean that the distance between A'B' and A''B'' is equal to the
distance between B'C' and B''C'', ( I'm indebted with El Filibustero
from it.scienza.matematica for this observation).
In particular there is one quadrilateral of this family, let's call
it A'B'C'D', that is circumscriptible (it can be constructed joining
the projections from the point of intersection of the diagonals of
SPQR to its sides; I learned this from P. Yiu's Euclidean Geometry
notes, pg. 165) .
Let's call r the radius of its incircle, O its center and h the
common distance between sides of A'B'C'D' and those of ABCD.
Let's suppose that, for example, BC+DA>=AB+CD (i.e. sides BC and DA
are the sides of ABCD that intersect the sides of A'B'C'D'):

A'B'=AB+ h tan(A/2)+h tan(B/2)
B'C'=BC - h tan(B/2)-h tan(C/2)
C'D'=CD + h tan(C/2)+h tan(D/2)
D'A'=DC - h tan(D/2) h tan(A/2)

but

A'B'+C'D'=B'C'+D'A'

So we can calculate h:

2h=(-AB+BC-CD+DA)/(tan (A/2)+ tan (C/2)+ tan (C/2)+ tan (D/2))

moreover, with easy trigonometric considerations:

2r=(A'B'+B'C'+C'D'+D'A')/(cot (A/2)+ cot (C/2)+ cot (C/2)+ cot (D/2))

but

A'B'+B'C'+C'D'+D'A'= AB+BC+CD+DA

so

2r=(AB+BC+CD+DA)/(cot (A/2)+ cot (C/2)+ cot (C/2)+ cot (D/2))

The area of ABCD can

2S(ABCD)=
=2(S(ABO)+S(CDO)+S(BCO)+S(DAO))=
=AB(r+h)+CD(r+h)+BC(r-h)+DA(r-h)=
=(AB+BC+CD+DA)r-(BC+DA-AB-CD)h

after substitution of h and r one obtains the sought formula.
I hope this all is correct.

Best regards
Ciao
Marcello Tarquini
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