There's a proof on how every obtuse angle is equal to 90 degrees, and I can't seem to find the issue.
Given: Quadrilateral ABCD, AD=BC, ∠ADB is obtuse, m∠CBD=90
Drawing perpendicular bisectors MP and NP from lines AB and CD (respectively) that intersect at the point P.
ΔPDA ≅ ΔPBC because of SSS
m∠PDB = m∠PBD because of base angles
m∠ADP = m∠CBP because of CPCTC
m∠ADB = m∠ADP – m∠PDB
m∠CBD = m∠CBP – m∠DBP
m∠ADB = m∠CBD
m∠ADB = 90
(the last few steps might be a little different depending on where point P is.)
Obviously, that's not true. I've already tried different cases, such that P was below, inside, and above the quadrilateral, but I can't seem to find the issue in this proof. I also know that P exists, since AC and BD aren't parallel.
Best Answer
You cannot have both:
Depending on the location of $P$, on will be
+and the other-.