Must the perp. bisectors of opposite sides in a quadrilateral intersect inside it, if one of the angles is right and an adjacent angle is obtuse?

Question

This is given that the lines aren't parallel so that the perpendicular bisectors will intersect at some point.

I've been trying to find a counterexample for this, but I can't, so I'm assuming that this statement is true, but I couldn't find any other information on the subject.

If it's true, how would I prove it?

EDIT: One of the angles is right and an adjacent angle is obtuse. This means the right angle and the obtuse angle are not opposite.

Answer 1

No, the statement is false. Construct any right trapezium, which is not a rectangle. The perpendicular bisectors of the 2 non-parallel sides will meet at the midpoint of the longer side. This can be proved fairly easily.

Can you try and prove it?

Answer 2

You can construct a counterexample; to show that it's false you need at least 1 example where it fails; for truth, you need to show a general statement where it holds:

Create two acute angles adjacent to that right angle; extend lines from those acute angles to close the quadrilateral. Draw perpendicular bisectors on either pair of sides; they will not intersect.

Try it. It's not rigorous at all, I know, but it works. On either side the perpendicular bisector away from the right angle will be deflected into the acute side.

Answer 3

Angle at B is $90^{\circ}.$ The shown perpendicular lines in the construction do not meet inside the quadrilateral, so the proposition is false.