One more exciting geometry problem

[Sujeet Kumar]

Q.In a triangle ABC, point P is taken so that angle ABP=70 degree, angle PBC=30 degree, angle BCP=10 degree and angle PCA=40 degree.Find angle BAP.

[Sujeet Kumar]

> Q.In a triangle ABC, point P is taken so that angle
> ABP=70 degree, angle PBC=30 degree, angle BCP=10
> degree and angle PCA=40 degree.Find angle BAP.

Here is the solution-

Q.In a triangle ABC, point P is taken so that angle ABP=70 degree, angle PBC=30 degree, angle BCP=10 degree and angle PCA=40 degree.Find angle BAP.

Sol.-Angle BAC=180-(70+30+10+40)=30 degree.
Let O be the circumcenter of triangle ABC.
Angle AOC(major)=2*(70+30)=200 degree.
Angle AOC(minor)=360-200=160 degree.
Triangle BPC wiil be congruent to triangle BPO.
Angle BOP=angle BCP=10 degree.
Angle POC=60-10=50 degree.
Angle AOP=angle AOC-angle POC=160-50=110 degree.
Considering quadrilateral ABPO,
angle ABP+angle AOP=70+110=180 degree.
So, quadrilateral ABPO is cyclic.
Angle BAP=angle BOP=10 degree.

Geometry Problem no.-4 is generalization of above problem.