Question 69
In the interior of an acute triangle $ABC$ the point $P$ is chosen, so that
$$
\angle APB = \angle CBA + \angle ACB, \quad \angle BPC = \angle ACB + \angle BAC.
$$
Prove that
$$
\frac{|AC| \cdot |BP|}{|BC|} = \frac{|BC| \cdot |AP|}{|AB|}. \qquad (\text{Belarus 2013})
$$
$$
\angle APB = \angle CBA + \angle ACB, \quad \angle BPC = \angle ACB + \angle BAC.
$$
Prove that
$$
\frac{|AC| \cdot |BP|}{|BC|} = \frac{|BC| \cdot |AP|}{|AB|}. \qquad (\text{Belarus 2013})
$$