1. **Problem 1:** Decide if the altitudes of a triangle always intersect inside the triangle. The altitudes of a triangle are the perpendicular lines from each vertex to the opposite side. - In an acute triangle, all altitudes intersect inside the triangle. - In a right triangle, the altitudes intersect at the right angle vertex, which is on the triangle. - In an obtuse triangle, the altitudes intersect outside the triangle. **Answer:** The altitudes intersect inside the triangle **sometimes true**. 2. **Problem 2:** Decide if in triangle ABC, $AB + BC > AC$. This is a statement of the triangle inequality theorem, which states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side. Therefore, $AB + BC > AC$ is **always true**. 3. **Problem 3:** Find the measure of $\angle ABC$ if $BD$ is an angle bisector. Given: - $BD$ bisects $\angle ABC$. - Segments on $BD$ are $x + 6$ and $2x - 3$. Since $BD$ is an angle bisector, it divides the opposite side $AC$ into segments proportional to the adjacent sides: $$\frac{AD}{DC} = \frac{AB}{BC}$$ But here, the problem gives lengths on $BD$, so we interpret $BD$ is split into $x+6$ and $2x-3$. Since $BD$ is an angle bisector, the two segments on $BD$ are equal: $$x + 6 = 2x - 3$$ Solve for $x$: $$x + 6 = 2x - 3$$ $$6 + 3 = 2x - x$$ $$9 = x$$ Now find $\angle ABC$: Since $BD$ bisects $\angle ABC$, the measure of $\angle ABC$ is twice the measure of one of the smaller angles. Calculate one smaller angle: $$x + 6 = 9 + 6 = 15$$ So $\angle ABC = 2 \times 15 = 30$ degrees. 4. **Problem 4:** Find $BC$ if $BD$ is a median of $\triangle ABC$. Given: - $BD$ is a median, so $D$ is midpoint of $AC$. - $AB = 3x - 5$ - $AC$ is divided into $5x - 5$ and $2x + 16$ at $D$. Since $D$ is midpoint of $AC$: $$5x - 5 = 2x + 16$$ Solve for $x$: $$5x - 5 = 2x + 16$$ $$5x - 2x = 16 + 5$$ $$3x = 21$$ $$x = 7$$ Now find $BC$: $$BC = 3x - 5 = 3(7) - 5 = 21 - 5 = 16$$ 5. **Problem 5:** Find $\angle 1$ if $BD$ is an altitude of $\triangle ABC$. Given: - $BD$ is altitude, so $BD \perp AC$. - Segments on $BD$ are $2x + 6$ and $x + 18$. Since $BD$ is altitude, $\angle 1$ is a right angle, so: $$m \angle 1 = 90^\circ$$ 6. **Problem 6:** Fill in the blanks for $\triangle SPU$. - a. Altitudes of $\triangle SPU$ are lines that are perpendicular from a vertex to the opposite side. - b. An angle bisector divides an angle into two equal parts. - c. A median connects a vertex to the midpoint of the opposite side. - d. A perpendicular bisector is a line that is perpendicular to a side and bisects it. From the figure (based on labels and right angles): - a. Altitudes: $UT$ and $VQ$ - b. Angle bisector: $PQ$ - c. Median: $VR$ - d. Perpendicular bisector: $QS$ **Summary:** - Problem 1: Sometimes true - Problem 2: Always true - Problem 3: $\angle ABC = 30^\circ$ - Problem 4: $BC = 16$ - Problem 5: $\angle 1 = 90^\circ$ - Problem 6: a. $UT$ & $VQ$, b. $PQ$, c. $VR$, d. $QS$