Dasappagupta\'s isoscles trapezium theorem

July 24, 2017 | Autor: R. Mudalagiri | Categoria: Gastroenterology, Geometry
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THEOREM No. 17
DASAPPAGUPTA'S ISOSCLES TRAPEZIUM THEOREM

In the isosceles trapezium if the parallel lines are in 1:4 ratio the diagonals are equal and also they trisect both base angles of the isosceles trapezium

Figure 43 Theorem 17

Data; GMDH is an isosceles trapezium with GH II MD and GH:MD is 1:4

Construction: MD line is divided into 4 equal parts MA=AR=RB=BD,
MD With A,R and B as centres and MA=AR=RB=BD as radius 3 circles are drawn. The
A circle and R circle intersect in G , and R circle and B circle intersect at H, Join lines
GH,MG,DH,GD and HM . GD and MH are two diagonals.
It is required to prove: HM=GD and they trisect both base angles of the isosceles trapezium .
Proof
In this construction Triangles AGR and GRH and HRB all are equal equilateral triangles and
They are all equal in all respects. Angles HGR=GRA=60degrees Therefore GH is parallel to AR and MD

Figure44 Theorem 17
The triangle HRD angle HRD=60 degree Angle RHD=90 degree and hence the remaining angle must be 30 degree. Angle GHR=Angle HRG=60 degree Therefore GH IIMD
Therefore both base angles of isosceles trapezium are equal to 30 degree.
In triangles HRM and GRD angles GRD=HRM=120 degrees by construction, GR=RH= radius of the same circle also MR=RD=1/2 MD
Therefore triangles HMR and GDR are equal in all respects. Therefore GD=HM, Therefore the diagonals are equal
GRD is an 120 degree obtuse angle triangle with 2GR=RD i.e 1:2 ratio The remaining 60 degrees has to be shared in 1:2 ratio. Angle RDG+ angle RGD= 60 degrees one should be 40 degree and the remaining must be 20 degree as per the obtuse angle triangle theorem
Since GH is parallel to MD angles HGD=GDR=20 degrees But we know angle HDR is 30 degree and angle RGH is 60 degrees.

Figure 45 Theorem 17
Therefore GD trisect the angle HDR into 10 and 20 degree and angle RGH into 40
and 20 degrees. Therefore the diagonals GD and HR trisect the angles on the both ends of the lines.
Now in triangles GMD and HDM , GM= HD MD is common and angles GMD=HDM=30
Degrees. Therefore both triangles are equal in all respects Therefore the lines GD and MH are
also equal Therefore the diagonals GD and MH of the Isosceles trapezium GMDH are
equal .Hence my father's new theorem came to existence from 1980 when he is on holiday in
UK in our house in Peterlee


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