Question 173
The point $P$ lies in the interior of triangle $ABC$ and satisfies
$$
\varangle B P C-\varangle B A C=\varangle C P A-\varangle C B A=\varangle A P B-\varangle A C B .
$$
Prove that then
$$
\overline{P A} \cdot \overline{B C}=\overline{P B} \cdot \overline{A C}=\overline{P C} \cdot \overline{A B}
$$
First one realizes that $\varangle B P C=60^{\circ}+\alpha, \varangle C P A=60^{\circ}+\beta, \varangle A P B=60^{\circ}+\gamma$. By symmetry it suffices to prove one of the two claimed equalities. (Hint: For an angle $\varangle X Y Z$ with $0^{\circ}\lt\varangle X Y Z\lt 180^{\circ}$ in the mathematically positive sense, set $\varangle Z Y X:= 180^{\circ}-\varangle X Y Z$. With this convention, for example, for four pairwise distinct points $W, X, Y, Z$ on a circle one always has $\varangle X Y Z=\varangle X W Z$).
$$
\varangle B P C-\varangle B A C=\varangle C P A-\varangle C B A=\varangle A P B-\varangle A C B .
$$
Prove that then
$$
\overline{P A} \cdot \overline{B C}=\overline{P B} \cdot \overline{A C}=\overline{P C} \cdot \overline{A B}
$$
First one realizes that $\varangle B P C=60^{\circ}+\alpha, \varangle C P A=60^{\circ}+\beta, \varangle A P B=60^{\circ}+\gamma$. By symmetry it suffices to prove one of the two claimed equalities. (Hint: For an angle $\varangle X Y Z$ with $0^{\circ}\lt\varangle X Y Z\lt 180^{\circ}$ in the mathematically positive sense, set $\varangle Z Y X:= 180^{\circ}-\varangle X Y Z$. With this convention, for example, for four pairwise distinct points $W, X, Y, Z$ on a circle one always has $\varangle X Y Z=\varangle X W Z$).