8.3. The height $CH$, dropped from the vertex of the right angle of triangle $ABC$, bisects the bisector $BL$ of this triangle. Find the angle $BAC$.

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The provided text has been translated from Russian to English, maintaining the original formatting and structure.
Answer: $30^{\circ}$.

Solution. Let $C H$ and $B L$ intersect at point $K$ (see Fig. 8.3). Then $C K$ is the median of the right triangle $B C L$, drawn to the hypotenuse, so $C K = 0.5 B L = B K$. Therefore, $\angle K C B = \angle K B C = \angle K B H$. Since the sum of these three angles is $90^{\circ}$ (from triangle $CBH$),

![](https://cdn.mathpix.com/cropped/2024_05_06_1c054e213dcb73e48b02g-1.jpg?height=269&width=574&top_left_y=2461&top_left_x=1301)
each of them is $30^{\circ}$. Consequently, $\angle C B A = 60^{\circ}, \angle B A C = 30^{\circ}$.

Grading Criteria.

「+」 A complete and well-reasoned solution is provided

「亡」 A generally correct reasoning is provided, containing minor gaps or inaccuracies that do not affect the answer

「Ғ」 It is proven that $CK = 0.5BL$, but further reasoning contains errors or is missing

「-」 Only the answer is provided

「-」 The problem is not solved or is solved incorrectly