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

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.

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.

Re: Help with this interesting proof (involving a trapezium)

1 Attachment(s)

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 $\displaystyle \angle OMN \neq \angle ODC$. Therefore if the triangles are similar, $\displaystyle \angle OMN = \angle OCD$, which implies that MN is parallel to AB and CD. Is this always true?