Not so tough geometry problem

[Sujeet Kumar]

Here is geometry problem which is not so tough-
Q.In a triangle ABC, point P is taken so that angle ABP=100 degree, angle PBC=20 degree and angle PCB=10 degree=angle PCA.Find angle BAP.

[Sujeet Kumar]

Here is the solution-

Q.In a triangle ABC, point P is taken so that angle ABP=100 degree, angle PBC=20 degree and angle PCB=10 degree=angle PCA.Find angle BAP.

Sol.-Take a point M on AC so that angle BMC=80 degree.
MC=BC.
Triangle PCM will be congruent to PCB.
MP=BP
Triangle MBP will be equilateral.
Angle ABM=120-80=40 degree.
Angle BAC=180-(100+20+10+10)=40 degree.
So, AM=BM=MP.
Triangle AMP is isosceles.
Angle AMP=180-20=160 degree.
So, angle MAP=10 degree.
Hence, BAP=40-10=30 degree.