Hi I have just found the proof that 90 equals 95 and was wondering if I have made some mistake. If so, which step in my proof is not true?
Definitions:
1. $\angle ABC=90^{\circ}$
2. $\angle BCD=95^{\circ}$
3. $|AB|=|CD|$
4. $M:=$ the center of $BC$
5. $N:=$ the center of $AD$
6. $l:=$ a line perpendicular to $BC$ passing through $M$
7. $m:=$ a line perpendicular to $AD$ passing through $N$
8. $S:=$ is the cross-section of $l$ and $m$
Based on definitions 1 through 8 we can draw the following image:

Based on the definitions we can derive the following:
9. $\triangle BSC$ is isosceles (follows from 4, 6 and 8)
10. $\triangle ASD$ is isosceles (follows from 5, 7 and 8)
11. $|BS|=|CS|$ (follows from 9)
12. $|AS|=|DS|$ (follows from 10)
13. $\triangle ABS\cong\triangle DCS$ (follows from 3, 11 and 12)
14. $\angle ABS=\angle DCS$ (follows from 13)
15. $\angle CBS=\angle BCS$ (follows from 9)
16. $\angle ABC=\angle ABS - \angle CBS=\angle DCS-\angle BCS=\angle BCD$ (follows from 14 and 15)
17. $90^{\circ}=95^{\circ}$ (follows from 1, 2 and 16)
Note: point $S$ is indeed lying above $BC$. If however it would be below $BC$, then the 'minus' in step 16 would simply have to be changed into a 'plus'.
Also note: The image is not drawn to scale. It only serves as to provide the reader with a intuitive view of the proof.
Also also note: before posting, I have first investigated if this type of question would be appropriate. Based on How come 32.5 = 31.5? and the following meta Questions about math jokes I have decided to post this question.
Your statement 16 depends on the way in which the diagram is drawn to know that ∠DCS = ∠DCB + ∠BCS.