Question 1208063
<pre>
<font color="red" size=4><b>Ikleyn, I'm going to show you your error.
</font></b>

{{{drawing(300,300,0,12,0,10,
line(1.4,2.6,11,9),line(11,1,11,9),line(4.85,4.9,11,6),line(5.15,1.975,4.85,4.9),line(1.4,2.6,11,1), locate(1.4,2.6,C), locate(5,2,Q),locate(12,0,B),
locate(11.15,6.2,P),locate(11.15,9.2,A),locate(4.5,5.3,R), locate(11,1,B),
line(11,9,5.15,1.975), locate(8.05,5.45,O),locate(8.55,6.16,alpha),
locate(7.05,5.3,alpha),locate(11.15,3.6,5),locate(11.15,7.7,3),
locate(4.55,3.55,3)


 )}}}

Let O be the intersection point of PR and AQ.

       Let  {{{alpha}}} be the angle AOP = QOR :  {{{alpha}}} = AOP = QOR
       (these angles are congruent since they are vertical angles).

<font color="red" size=4><b>So far, so good.</font></b>

Since the areas of triangle ABQ and quadrilateral PBQR are equal, we conclude from this fact 
that the areas of triangles AOP and ROQ are equal. The equation for these equal areas is

    {{{(1/2)*OA*OP*sin(alpha)}}} = {{{(1/2)*OQ*OR*sin(alpha)}}}.    (1)

<font color="red" size=4><b>So far, so good. That's the SAS area formula.</font></b>

{{{drawing(200,1200/7,-1,6,-1,5,
locate(2.9,1.9,O),
line(1,0,1,3),line(1,3,5,1),line(5,1,5,4),line(1,0,5,4),
locate(5,1,P),locate(5,4.5,A),locate(1,3.5,R), locate(.9,0,Q),

locate(3.3,2.4,alpha),locate(2.3,2.2,alpha),


locate(5.1,2.7,3),locate(.625,1.7,3)

)}}}{{{drawing(200,1200/7,-1,6,-1,5,

line(1.41886116991581033400,0.418861169915810334001,5,4),line(0.4701778718652965344,3.26491106406735173280,5,1),line(5,1,5,4),line(1.41886116991581033400,0.418861169915810334001,0.4701778718652965344,3.26491106406735173280),
locate(2.9,1.9,O),locate(5.1,2.7,3), locate(.625,1.9,3),

locate(3.3,2.4,alpha),locate(2.3,2.2,alpha),

locate(5,1,P),locate(5,4.5,A), locate(.4,3.8,R), locate(1.3,.44,Q)
)}}}

After reducing, from (1) we have 

    OA*OP = OQ*OR.    (2)

<font color="red" size=4><b>So far, so good!  But be aware that is true for both cases above, not for just
the one on the left.</font></b> 

It leads to proportion

    {{{abs(OA)/abs(OP)}}} = {{{abs(OR)/abs(OQ)}}}.    (3)

<font color="red" size=4><b>So far, so good!  But I repeat, that is also true for both cases above, not for
just the one on the left.</font></b>

Thus we see that triangles  AOP  and  QOR  have congruent angles  AOP  and  QOR  and proportional sides
that conclude these angles.  So, these triangles  AOP  and  QOR  are  SIMILAR.

<font color="red" size=4><b>So far, so good.  But I'm still warning you, that is also true for both cases.
above, not for just the one on the left.</font></b>
 
So, the triangles  AOP  and  QOR  are similar.
From it, we conclude that these triangles have congruent corresponding angles.

<font color="red" size=4><b>That is also true for both cases above, not for just the one on the left.</font></b>

In particular, angles  OAP  and  OQR  are congruent.

<font color="red" size=4><b>AHA!!! There is your fallacy right there!!!  You cannot assume that!!!!! 
It could be angles OAP and ORQ that are congruent.

See, Ikleyn?  We all make mistakes.  Are you saying that this is not a mistake?
You know very well it is!  I apologize for getting angry. But proofs are either
valid or not.  If a proof is not valid, no amount of whining or trashing me will
make it valid.  You must prove that AP and QR are parallel.  You have not done that.  
Why not admit that you can make mistakes?  
  
Edwin</font></b></pre>