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Re: Parallelogram Orthocenters
Posted:
Oct 16, 2013 12:24 PM


> Hi Avni, > > I suppose locating orthocenter by finding altitudes ( > = 2 Area of triangles / base ) etc. can be done. > > By geometric construction H1H2 ~ 7.57188, its > component parallel to AB is 7. It can be noted that > O, midpoint of DB or H1H2 is the antisymmetric > centre of opposite vertices of parallelogram ABCD. > > Regards > Narasimham
Hi Narasimham,
you are right, but of course an analytical solution is required. I gave some numerical values of side lengths and angles only for clarity, but they are fully unimportant. Let it be AB=DC=a, AD=BC=b, and angle(BAD)= fi. There are some further interesting properties of this structure that I want to reveal after an analytical solution is provided.
Best regards, Avni



