Question

which of the following statements are true? check all of the boxes that apply. triangle abc is a right triangle. segment h divides triangle abc into two congruent triangles segment h divides triangle abc into two right triangles. a² = b² + c² c² = h² + b² - 2bx + x² if you know sides a, b, and c, you can find all the angles of triangle abc. done

Explanation:

Step 1: Analyze the triangle type

Since there is no indication that triangle \(ABC\) has a right angle (the right - angle symbol is at the intersection of \(h\) and \(b\), not at a vertex of \(\triangle ABC\)), the statement “Triangle \(ABC\) is a right triangle” is false.

Step 2: Check congruent triangles

For two triangles to be congruent, all corresponding sides and angles must be equal. Here, we don't know if \(a = c\) (if \(a
eq c\), the two sub - triangles formed by \(h\) will not be congruent as one side \(a\) and \(c\) is different). So, the statement “Segment \(h\) divides triangle \(ABC\) into two congruent triangles” is false.

Step 3: Check right - angled sub - triangles

Since \(h\perp b\), the two sub - triangles (the one with sides \(x\) and \(h\) and the one with sides \(b - x\) and \(h\)) are right - angled. So, the statement “Segment \(h\) divides triangle \(ABC\) into two right triangles” is true.

Step 4: Check \(a^{2}=b^{2}+c^{2}\)

By the Pythagorean theorem, \(a^{2}=h^{2}+x^{2}\) and \(c^{2}=h^{2}+(b - x)^{2}\). Since \(a^{2}
eq b^{2}+c^{2}\) (because \(b^{2}+c^{2}=b^{2}+h^{2}+(b - x)^{2}=2b^{2}-2bx+x^{2}+h^{2}
eq h^{2}+x^{2}\) in general), the statement “\(a^{2}=b^{2}+c^{2}\)” is false.

Step 5: Derive \(c^{2}=h^{2}+b^{2}-2bx + x^{2}\)

Using the Pythagorean theorem in the right - triangle with sides \(h\), \(b - x\), and \(c\), we have \(c^{2}=h^{2}+(b - x)^{2}\). Expanding \((b - x)^{2}=b^{2}-2bx+x^{2}\), so \(c^{2}=h^{2}+b^{2}-2bx + x^{2}\). The statement “\(c^{2}=h^{2}+b^{2}-2bx + x^{2}\)” is true.

Step 6: Check angle determination

Using the Law of Cosines (\(\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}\), \(\cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}\), \(\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}\)), if we know \(a\), \(b\), and \(c\), we can find all angles of \(\triangle ABC\). The statement “If you know sides \(a\), \(b\), and \(c\), you can find all the angles of triangle \(ABC\)” is true.

Answer:

Segment \(h\) divides triangle \(ABC\) into two right triangles, \(c^{2}=h^{2}+b^{2}-2bx + x^{2}\), If you know sides \(a\), \(b\), and \(c\), you can find all the angles of triangle \(ABC\).