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Re: Parallelogram Orthocenters
Posted:
Oct 16, 2013 12:06 PM


> Let ABCD be a parallelogram such that AB=DC=10, and > AD=BC=6, > and angle(BAD)=60°. Let H1,H2 be the orthocenters of > triangles ABD and CDB respectively. Determine the > distance between H1 and H2. > > Best regards, > Avni
A pretty figure. Two normals divide the parallelogram sides into 3+7 and 1+5 so that the orthocenters are the opposite corners of a rectangle 7x(3.sqrt32.2/sqrt3) = 7 x (5/sqrt3) whence H1H2 = sqrt(172/3) = 2.sqrt(43/3)
Regards, Peter Scales.



