Proof that obtuse angles = 90 degrees

Question

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

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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.

Answer 1

You cannot have both:

m∠CBD = m∠CBP - m∠DBP
m∠CBD = m∠CBP - m∠DBP

Depending on the location of $P$, on will be + and the other -.

Answer 2

"ΔPDA ≅ ΔPBC because of SSS"

No. ΔPDA ≅ ΔPCB because PD = PC (not PB) and PA = PB (not PC)

So you have the corresponding angles flipped.

"m∠PDB = m∠PBD because of base angles"

So No. "m∠PDC = m∠PCD" and so on. The whole thing falls apart.

(I got confused because I drew the picture of the quad as ADBC rather than ABCD and I drew the perpendiculars of DB and AC, rather than of AB and DC.)

Draw the picture carefully and label the points correctly.