Question 68
The convex quadrilateral $ABCD$ is given.
$\angle BAD = \angle BCD = 120^\circ$, $\angle ACD = 80^\circ$, $BC = CD$, $O$ is the intersection point of its diagonals.
Prove, that $BO^2 = AO \cdot AC$.
$\angle BAD = \angle BCD = 120^\circ$, $\angle ACD = 80^\circ$, $BC = CD$, $O$ is the intersection point of its diagonals.
Prove, that $BO^2 = AO \cdot AC$.