Geometry problem no.-6

[Sujeet Kumar]

Here is geometry problem no.-6:-
Q.In a triangle ABC,point P is taken so that angle ABC=30 degree=angle PCB, angle PBC=x degree and angle PCA=(60-2x) degree.Find angle BAP.

**Please! Please!! Please !!! Any body reply.Any comments, +ve or -ve .If I know I am doing something which is totally worthless, then I will stop.**

[Philippe 92]

Dear Sujeet,

The lack of answers doesn't mean your problems are uninterresting,
but just that nobody seems to have found a solution !
(however just few people read this forum)

For instance for this problem, we may consider the fixed triangle
BCD with 30-60-90 angles, and the fixed midpoint E of BD, with
obviously CDE equilateral.

Then x is defined by a free point P on CE, and A is deduced by
angle DCA = 2x
(See attachement)

But then ???
How could we prove that APCI is cyclic and BAP is therefore
BAP = 60 - 2x + x = 60 - x ???

Best Regards.

Attachment available from http://mathforum.org/kb/servlet/JiveServlet/download/129-2225675-7351334-663457/sol166-1.gif

[Sujeet Kumar]

> Dear Sujeet,
>
> The lack of answers doesn't mean your problems are
> uninterresting,
> but just that nobody seems to have found a solution
> !
> (however just few people read this forum)
>
> For instance for this problem, we may consider the
> fixed triangle
> BCD with 30-60-90 angles, and the fixed midpoint E of
> BD, with
> obviously CDE equilateral.
>
> Then x is defined by a free point P on CE, and A is
> deduced by
> angle DCA = 2x
> (See attachement)
>
> But then ???
> How could we prove that APCI is cyclic and BAP is
> therefore
> BAP = 60 - 2x + x = 60 - x ???
>
> Best Regards.

Directly I can't prove that APCI is cyclic.But I am little bit confused , you are asking or hinting at something interesting.
Besr Regards , Sujeet