Not a new geometry problem

[Sujeet Kumar]

Here is a geometry problem , which is not very much different from old ones-
Q.In a triangle ABC, point P is taken so that angle ABP=30 degree, angle PBC=50 degree,angle PCB=30 degree and angle PCA=10 degree.Find angle BAP.

[Sujeet Kumar]

Here is the solution-

Q.In a triangle ABC, point P is taken so that angle ABP=30 degree, angle PBC=50 degree,angle PCB=30 degree and angle PCA=10 degree.Find angle BAP.

Sol.-Take a point M on the same side of BC as P, so that triangle MBC is equilateral.
Angle BAC=180-(30+50+30+10)=60 degree=angle BMC.
So, quadrilateral AMCB is cyclic.
Angle MAC=angle MBC=60 degree.
Angle AMB=angle ACB=40 degree.
Triangle PCB will be congruent to triangle PCM.
Angle PMC=angle PBC=50 degree.
Angle BMP=60-50=10 degree.
Let BM and AC intersect at Q.
Considering quadrilateral PQMC,
angle QMP=10 degree=angle QCP
So, quadrilateral PQMC will be cyclic.
Angle QPM=angle QCM=60-40=20 degree.
Considering triangle AMP and placement of point Q.
This configuration corresponds to our earlier solved problems(Boring Geometry problem).