SOLUTION: The area of △ABC is 40. Points P, Q and R lie on sides AB, BC and CA respectively. If AP = 3 and PB = 5, and the area of △ABQ is equal to the area of PBQR, determine the area o

Before to move forward, make a sketch of the problem. 
I will assume that the sketch is in front of your eyes.


       Let O be the intersection point of PR and AQ.

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


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)


After reducing, from (1) we have 

    OA*OP = OQ*OR.    (2)


It leads to proportion

     = .    (3)


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


    It is the key idea of the solution.
    What follows, is the direct consequence of this idea.



So, the triangles  AOR  and  QOP  are similar.
From it, we conclude that these triangles have congruent corresponding angles.

In particular, angles  OAR  and  OQP  are congruent.

It implies that line  AR  is parallel to   PQ.


It is the same as to say that  PQ  is parallel to  AC.


Hence, triangle  PBQ  is similar to triangle  ABC.



From this similarity, we have   =  =  = .


Thus we proved that under given condition, point Q divides the side BC in the same proportion  
as the point P divides side AB.


Hence, the area of triangle  AQC  is    of the area of triangle  ABC,  i.e.   = 3*5 = 15 square units.


At this point, the problem is solved completely.

There is a fallacy in Ikleyn's proof above.  I will draw just enough of it to
show the fallacy.  While she is correct that triangles AOP and QOR are
congruent, QR is not necessarily parallel to AP, and therefore not necessarily
parallel to AB as she has stated. I do not see that she has proved it parallel.

We must rule out the possibility of the figure on the right below, and show that
the figure on the left is the correct one. She is correct that QR = AP = 3, but
she must show that QR is parallel to AP.  I think she can show this by making
use of the fact that QR = 3 and the converse of this well-known theorem:

If a line parallel to one side of a triangle intersects the other two sides, 
then it divides the two sides proportionally. (The converse is also true.)

If Ikleyn will correct this error in her proof, I will delete this post. So that
the student will learn of the corrected proof, I will make another post in
one of my other screen names, to inform him or her of the correction.  



Below I have drawn the figure for the whole problem, but I purposely drew QR to
look not quite parallel to AB, which Ikleyn will prove it to be. She is welcome
to use my figure below.  



Edwin

After thinking about it, maybe it isn't necessary to prove QR is parallel to AP,
but only that the areas of those two triangles are equal.  Each would add the
same amount of area to quadrilateral PBQO.  I just noticed that QR was not proved
parallel to AP, though it was stated that they were parallel.  It would also be
interesting to prove that they are parallel.  That might be more of a challenge
than the original problem. 

Edwin

But after thinking about it some more, it is necessary to prove QR is parallel
to AP, for if QR and AP weren't parallel, then triangles RQC and ABC would not be
similar. So while it wouldn't matter to get the area of triangle ABQ equal to the
area of quadrilateral PBQR, it is absolutely necessary to determine the area of
triangle AQC.  We'll see what Ikleyn can do about proving them parallel. I now
think it's necessary.

Edwin

Ikleyn is too much like Trump.  She can't stand to be wrong.  But she is. 
She has not proved QR is parallel to AB. And that is absolutely necessary to
determine the area of AQC.  Sure, she got the right answer, but the question
is to prove it, not get it!

She did not show that angles OAP and QAR are congruent. Just because the product
of two sides are equal, does not show which way the triangles are turned.

Edwin

Ikleyn, I'm going to show you your error.




Let O be the intersection point of PR and AQ.

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

So far, so good.

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)

So far, so good. That's the SAS area formula.



After reducing, from (1) we have 

    OA*OP = OQ*OR.    (2)

So far, so good!  But be aware that is true for both cases above, not for just
the one on the left. 

It leads to proportion

     = .    (3)

So far, so good!  But I repeat, that is also true for both cases above, not for
just the one on the left.

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.

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.
 
So, the triangles  AOP  and  QOR  are similar.
From it, we conclude that these triangles have congruent corresponding angles.

That is also true for both cases above, not for just the one on the left.

In particular, angles  OAP  and  OQR  are congruent.

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