Problem 8.7. In an acute-angled triangle $A B C$, the altitude $B H$ is drawn. It turns out that $C H=A B+A H$. How many degrees does the angle $B A C$ measure if $\angle A B C=84^{\circ}$ ?

![](https://cdn.mathpix.com/cropped/2024_05_06_b135fe79f2e6ed0cbffag-4.jpg?height=367&width=466&top_left_y=1395&top_left_x=502)
Answer: 64.

![](https://cdn.mathpix.com/cropped/2024_05_06_b135fe79f2e6ed0cbffag-5.jpg?height=311&width=505&top_left_y=93&top_left_x=468)

Fig. 8: to the solution of problem 8.7

Solution. Mark a point \( K \) on the segment \( CH \) such that \( AH = HK \). Then, from the condition, it follows that \( AB = CK \) (Fig. 8).

In triangle \( ABK \), the height \( BH \) coincides with the median, so it is isosceles, \( AB = BK \), and \( \angle BAH = \angle BKA \).

Let \( \angle ACB = x \). Since \( CK = AB = BK \), triangle \( BCK \) is isosceles, and \( \angle KBC = \angle KCB = x \). Then \( \angle AKB = 2x \), \( \angle KAB = 2x \), and \( \angle ABC = 180^\circ - 3x \).

Since \( 84^\circ = \angle ABC = 180^\circ - 3x \), we find \( x = 32^\circ \). Therefore, \( \angle BAC = 2 \cdot 32^\circ = 64^\circ \).