Geometry problem no.-4

[Sujeet Kumar]

Here is geometry problem no.-4 :-
Q.In a triangle ABC , point P is taken so that angle ABP=(60+x) degree,angle PBC=30 degree, angle PCB=x degree and angle PCA=(60-2x) degree.Find angle BAP.

As, when x=10 or multiples of 10 degree, then sides and cevian can be seen as sides and diagonals of
18 gon.In the same way with suitable replacement of x, sides and cevians of this problem will become
diagonals and sides of higher gons.For example 30-gon, 24 -gon , 60-gon or 360-gon.
I am very bad in expressing myself, but this time I hope, I have made myself clear.

[Sujeet Kumar]

Here is the solution:-

Q.In a triangle ABC , point P is taken so that angle ABP=(60+x) degree,angle PBC=30 degree, angle PCB=x degree and angle PCA=(60-2x) degree.Find angle BAP.

Sol.-Angle BAC=180-(30+60+x+x+60-2x)=180-150=30 degree.
Let O be the circumcenter of triangle ABC.
Angle AOC(major)=2*(60+x+30)=(180+2x) degree
Angle AOC(minor)=360-(180+2x)=(180-2x) degree.
Triangle BPC will be congruent to BPO.
Angle BOP=angle BCP=x degree.
Angle POC=(60-x) degree.
Angle AOP=angle AOC-angle POC=(180-2x)-(60-x)=(120-x) degree.
Considering quadrilateral ABPO,
Angle ABP+angle AOP=60+x+120-x=180 degree.
So, quadrilateral ABPO is cyclic.
Angle BAP=angle BOP= x degree.