# Help with this interesting proof (involving a trapezium)

• Aug 4th 2012, 02:04 AM
ThetaMaster11
Help with this interesting proof (involving a trapezium)
Let ABCD be a trapezoid, with AB parallel to CD (the vertices are listed in cyclic order). The
diagonals of this trapezoid are perpendicular to one another and intersect at O. The base
angles DAB and CBA are both acute. A point M on the line segment OA is such
that BMD = 90degrees, and a point N on the line segment OB is such that ANC = 90degrees.
Prove that triangles OMN and OBA are similar.

Cheers
• Aug 7th 2012, 11:21 PM
ksoaknight51
Re: Help with this interesting proof (involving a trapezium)
i dont understand how the base angles can be acute, and also ANC and BMD cannot equal 90. You might want to try Art of problem solving website, for more help.
• Aug 7th 2012, 11:50 PM
richard1234
Re: Help with this interesting proof (involving a trapezium)
Nothing wrong with the problem. I was able to draw a diagram that satisfies all the conditions.

ABO is similar to CDO, so you want to prove that triangles MNO and CDO are similar. There are probably multiple solutions...I see one involving cyclic quadrilaterals.
• Aug 8th 2012, 01:57 AM
ksoaknight51
Re: Help with this interesting proof (involving a trapezium)
anyone have a solution?
• Aug 8th 2012, 10:09 AM
richard1234
Re: Help with this interesting proof (involving a trapezium)
Just realized my cyclic quadrilateral solution doesn't work. I'm under the impression that you can't prove those two triangles are similar.

Attachment 24466

Clearly, quadrilateral MNCD is not cyclic, so $\angle OMN \neq \angle ODC$. Therefore if the triangles are similar, $\angle OMN = \angle OCD$, which implies that MN is parallel to AB and CD. Is this always true?