right angle equal to obtuse triangle?

Question

Given the obtuse angle x, we make a quadrilateral $ABCD$ with $∠DAB = x$, and $∠ABC = 90◦$, and $AD = BC$. Say the perpendicular bisector to $DC$ meets the perpendicular bisector to $AB$ at $Q$. Then $QA = QB$ and $QC = QD.$ So the triangles $QAD$ and $QBC$ have equal sides and are congruent. Thus $∠QAD = ∠QBC.$ But $QAB$ is isosceles, hence $∠QAB = ∠QBA.$ Subtracting, gives $x = ∠QAD−∠QAB = ∠QBC −∠QBA = 90◦$

When I draw out the actual figure, I find out that the proof falls at the last part; $QD$ will pass below A so $∠QAD$ does not equal to $x+∠QAB$ but I can't provide a formal prove of it. Some suggestion will be appreciated

Answer 1

You argument is the proof you are looking for. If $Q$ were situated such that $\angle QAD=\angle DAB+\angle QAB$, then your argument would be valid, and prove that $x$ is a right angle. Since this conclusion is false, this is a proof by contradiction that $Q$ cannot be so situated.

Answer 2

You have proven by contradiction that $Q$ can NOT be in the interior of the quadrilateral becasuse if it were you would have proven the incorrect result that $x$ is right.

You'd get a similar result if $\angle DAB$ were accute.

If $\angle BAD$ is right then the quadrilateral is a recatangle and the bisectors are the same line and any interior point will do.

Otherwise if $\angle BAD$ is not right then the lines $DC$ and lines $AB$ are not parallel. So the perp bisectors intersect. You've proven they can't meet on the interior. And the other hand you haven't proven whether the intersect above or below.