Let's go through the next steps in detail after the initial steps provided:
1. Draw an altitude from [tex]\(B\)[/tex] to segment [tex]\(\overline{AC}\)[/tex]. Label the point where the altitude and segment [tex]\(\overline{AC}\)[/tex] intersect as [tex]\(D\)[/tex]. The altitude forms two right triangles within [tex]\(\triangle ABC\)[/tex].
2. By the Pythagorean theorem:
- For [tex]\(\triangle ABD\)[/tex]: [tex]\(c^2 = h^2 + x^2\)[/tex]
- For [tex]\(\triangle BDC\)[/tex]: [tex]\(a^2 = h^2 + (b - x)^2\)[/tex]
3. Expanding the binomial squared:
[tex]\[
a^2 = h^2 + b^2 - 2bx + x^2
\][/tex]
4. Subtract the equation for [tex]\(c^2\)[/tex] from [tex]\(a^2\)[/tex]:
[tex]\[
a^2 - c^2 = (h^2 + b^2 - 2bx + x^2) - (h^2 + x^2)
\][/tex]
Simplify:
[tex]\[
a^2 - c^2 = b^2 - 2bx
\][/tex]
5. Subtract [tex]\(b^2\)[/tex] from both sides of the equation:
[tex]\[
a^2 - b^2 - c^2 = -2bx
\][/tex]
6. Add [tex]\(2bx\)[/tex] to both sides of the equation:
[tex]\[
a^2 - b^2 - c^2 + 2bx = 0
\][/tex]
Simplify:
[tex]\[
a^2 - c^2 + 2bx = b^2
\][/tex]
7. Add [tex]\(c^2\)[/tex] to both sides of the equation:
[tex]\[
a^2 = b^2 + c^2 - 2bx
\][/tex]
To continue the proof and simplify this expression further, consider the relationship between [tex]\(x\)[/tex] and [tex]\(\cos(A)\)[/tex]:
8. By construction, [tex]\(x = c \cos(A)\)[/tex]. Substitute this into the equation:
[tex]\[
a^2 = b^2 + c^2 - 2b \cdot (c \cos(A))
\][/tex]
9. Therefore, the proof of the law of cosines is completed:
[tex]\[
a^2 = b^2 + c^2 - 2bc \cos(A)
\][/tex]
And thus, we have successfully proven the law of cosines for a triangle with any angles, including acute angles.
