Triangle Relationships — Free Geometry Review Games.

A midsegment connects the midpoints of two sides of a triangle.

The triangle midsegment theorem says it is half the length of the third side.

3 + 4 = 7 < 8. The triangle inequality fails, so no triangle exists.

The centroid is the intersection of the three medians.

In any triangle, the longest side is across from the largest angle.

5+7=12>11, 5+11>7, 7+11>5. All inequalities hold, so yes.

The centroid is 2/3 of the way from each vertex to the opposite midpoint.

The circumcenter is equidistant from all three vertices.

The incenter is equidistant from all three sides.

The largest angle is opposite the longest side (10).

Centroid divides median 2:1 from vertex: 6 (vertex to centroid) and 3 (centroid to midpoint).

Shorter sides are opposite smaller angles.

The circumcenter is the intersection of the perpendicular bisectors.

The orthocenter is the intersection of the three altitudes.

1 + 2 = 3 < 4. The triangle inequality fails.

By the Midsegment Theorem, the segment connecting the midpoints of two sides of a triangle is parallel to the third side and equals half its length. Since \(M\) and \(N\) are midpoints of \(\overline{AB}\) and \(\overline{AC}\), \(MN = \frac{1}{2} \cdot BC = \frac{1}{2} \cdot 20 = 10\). Choice B is wrong because the midsegment equals half the length of the parallel side, not the full length.

The Triangle Inequality Theorem requires that the sum of any two sides must be strictly greater than the third side. Here, \(1 + 1 = 2\), which is not greater than \(3\), so no valid triangle can be formed. Choice D is wrong because having positive side lengths is necessary but not sufficient — the inequality condition must also be satisfied.

A median goes from one vertex to the midpoint of the opposite side. Choice B describes a midsegment. Choice C describes an altitude. Choice D describes an angle bisector. Each of these four segments has a distinct definition and different special properties.

The incenter is the intersection of the three angle bisectors and is the center of the inscribed circle (incircle). It is equidistant from all three sides. The centroid is where medians meet; the circumcenter is where perpendicular bisectors meet; and the orthocenter is where altitudes meet.

By the Triangle Inequality, the third side must satisfy \(|9 - 5| < x < 9 + 5\), giving \(4 < x < 14\). Choice B is wrong because it allows values like \(x = 1\), which would make \(5 + 1 = 6\), not greater than \(9\). The inequalities are strict (not \(\leq\)), so Choice D is also incorrect.

An angle bisector divides one interior angle of the triangle into two equal angles. Choice B describes a median. Choice C describes an altitude. Choice D describes a midsegment. These four types of special segments are commonly confused, so knowing each definition precisely is essential.

Every triangle has exactly \(3\) medians — one drawn from each vertex to the midpoint of the opposite side. All three medians of any triangle intersect at a single point called the centroid. Since a triangle has three vertices and three opposite sides, it has exactly three medians.

By the Midsegment Theorem, \(ST = \frac{1}{2} \cdot QR\), so \(3x - 2 = \frac{4x + 6}{2} = 2x + 3\). Solving gives \(x = 5\). Verify: \(ST = 13\) and \(QR = 26\), and indeed \(13 = \frac{26}{2}\). Choice B (\(x = 3\)) gives \(ST = 7\) and \(QR = 18\), but \(7 \neq \frac{18}{2}\), so it fails the check.

The side opposite a larger angle is longer. The smallest angle is \(\angle A = 45°\), opposite \(BC\); the middle angle is \(\angle C = 60°\), opposite \(AB\); the largest is \(\angle B = 75°\), opposite \(AC\). So the order is \(BC < AB < AC\). Choice B incorrectly swaps \(AB\) and \(AC\), reversing the relationship between the \(60°\) and \(75°\) angles.

The centroid divides each median in the ratio \(2:1\) from the vertex. So \(AG : GM = 2 : 1\), meaning \(GM = \frac{AG}{2} = \frac{8}{2} = 4\). The full median has length \(AM = 12\). Choice B (\(GM = 8\)) would imply a \(1:1\) ratio, which is incorrect — the centroid is closer to the midpoint, not at the middle of the median.

By the Angle Bisector Theorem, \(\frac{YD}{DZ} = \frac{XY}{XZ} = \frac{9}{6} = \frac{3}{2}\). Since \(YD + DZ = 10\), we get \(YD = \frac{3}{3+2} \cdot 10 = 6\) and \(DZ = 4\). Choice B (\(4\)) is actually the length of \(DZ\), not \(YD\) — a common error from assigning the ratio to the wrong segment.

For \(\{3, 5, 9\}\): \(3 + 5 = 8\), which is not greater than \(9\), violating the Triangle Inequality. The other three sets all pass: \(7 + 8 = 15 > 12\); \(5 + 6 = 11 > 10\); \(4 + 9 = 13 > 11\). Only the smallest two sides need to be checked against the largest side.

When sides are ordered \(a < b < c\), the inequality \(a + b > c\) is the binding (hardest to satisfy) constraint. If it holds, the other two automatically hold: since \(c > b\), we have \(a + c > a + b > c > b\), so \(a + c > b\); similarly \(b + c > a\) is trivially true. Choices B and C are always satisfied for positive lengths once \(c\) is the largest side.

By the Midsegment Theorem, \(MN \parallel AC\) and \(MN = \frac{1}{2} AC\). Therefore \(AC = 2 \cdot MN = 2 \cdot 5 = 10\). The perimeter of triangle $MBN$ is extra information — a distractor not needed to find \(AC\). Choice B (\(5\)) confuses \(MN\) with \(AC\) itself.

The Midsegment Theorem states the segment connecting midpoints of two sides is parallel to the third side and equal to half its length. Since \(P\) and \(Q\) are midpoints of \(\overline{DE}\) and \(\overline{DF}\), the midsegment \(\overline{PQ}\) is parallel to \(\overline{EF}\) (the side not used) with \(PQ = \frac{1}{2}EF\). Choice D incorrectly identifies the parallel side as \(\overline{DE}\), which is one of the sides being bisected, not the third side.

The largest angle is always opposite the longest side. Since \(\angle A = 110°\), the other two angles must sum to \(70°\) and are each less than \(110°\). The side opposite \(\angle A\) is \(BC\), so \(BC\) must be the longest side. Choices B and C are wrong because \(AB\) and \(AC\) are adjacent to \(\angle A\), not opposite it. Choice D is impossible since equal sides require equal angles.

By the Angle Bisector Theorem, \(\frac{AD}{DC} = \frac{AB}{BC}\), so \(\frac{AB}{BC} = \frac{4}{6} = \frac{2}{3}\). Choice B (\(\frac{3}{2}\)) inverts the ratio — a common error. The theorem matches each segment of the divided side with the adjacent triangle side on the same side of the bisector.

Being equidistant from all three vertices is a property of the circumcenter, not the incenter. The incenter is equidistant from all three sides (that distance is the inradius). Choices B, C, and D are all true for the incenter. Unlike the circumcenter and orthocenter, the incenter always lies inside the triangle.

By the Midsegment Theorem, \(MN = \frac{1}{2} BC\), so \(2x + 4 = \frac{5x - 3}{2}\). Multiplying both sides by \(2\) gives \(4x + 8 = 5x - 3\), so \(x = 11\). Verify: \(MN = 26\) and \(BC = 52\), and \(26 = \frac{52}{2}\) ✓. Choice B (\(x = 5\)) gives \(MN = 14\) and \(BC = 22\), but \(14 \neq \frac{22}{2}\).

The centroid divides each median in the ratio \(2:1\) from the vertex, so the centroid lies \(\frac{1}{3}\) of the total length from the midpoint. Thus \(GA = \frac{1}{3} PA\), giving \(PA = 3 \cdot 4 = 12\); and \(GB = \frac{1}{3} QB\), giving \(QB = 3 \cdot 6 = 18\). Choice B doubles instead of triples, confusing the vertex-to-centroid distance (\(\frac{2}{3}\) of the median) with the midpoint-to-centroid distance (\(\frac{1}{3}\)).

By the Angle Bisector Theorem, \(\frac{BD}{DC} = \frac{AB}{AC} = \frac{10}{8} = \frac{5}{4}\). Since \(BD + DC = 6\), we get \(BD = \frac{5}{5+4} \cdot 6 = \frac{5}{9} \cdot 6 = \frac{10}{3}\). Choice C (\(BD = 3\)) would mean \(DC = 3\) giving a \(1:1\) ratio, which would require \(AB = AC\), but \(10 \neq 8\).

Apply all three triangle inequalities. The inequality \(x + (x + 2) > 2x - 1\) simplifies to \(2 > -1\), always true. The inequality \((x+2) + (2x-1) > x\) simplifies to \(3x + 1 > x\), always true for \(x > 0\). The binding constraint is \(x + (2x - 1) > x + 2\), which gives \(3x - 1 > x + 2\), so \(2x > 3\) and \(x > \frac{3}{2}\). We also need \(2x - 1 > 0\), giving \(x > \frac{1}{2}\), but \(\frac{1}{2} < \frac{3}{2}\) so \(x > \frac{3}{2}\) is the binding answer.

The centroid divides each median \(\frac{2}{3}\) of the way from each vertex: \(AG = \frac{2}{3}(15) = 10\), \(BG = \frac{2}{3}(12) = 8\), \(CG = \frac{2}{3}(9) = 6\). So \(AG + BG + CG = 10 + 8 + 6 = 24\). Choice B (\(18\)) results from using \(\frac{1}{3}\) of each median instead — those would be the distances from each midpoint to the centroid (\(GD + GE + GF\)), not from the vertices.

Note that \(\sqrt{20} = 2\sqrt{5}\). The third side \(c\) must satisfy \(|2\sqrt{5} - \sqrt{5}| < c < 2\sqrt{5} + \sqrt{5}\), giving \(\sqrt{5} < c < 3\sqrt{5}\). Since \(\sqrt{5} \approx 2.24\) and \(3\sqrt{5} \approx 6.71\), choice A (\(c = 3\)) satisfies \(2.24 < 3 < 6.71\). Choice B (\(c = \sqrt{5}\)) equals the lower bound exactly (not strictly greater). Choice C (\(c = 3\sqrt{5}\)) equals the upper bound exactly. Choice D (\(c = \frac{\sqrt{5}}{2} \approx 1.12\)) is below the lower bound.

Since \(\angle A + \angle B + \angle C = 180°\), substituting gives \(\angle A + 2\angle A + 3\angle A = 6\angle A = 180°\), so \(\angle A = 30°\), \(\angle B = 60°\), \(\angle C = 90°\). The side opposite the largest angle (\(\angle C = 90°\)) is \(AB\); opposite \(\angle B = 60°\) is \(AC\); opposite \(\angle A = 30°\) is \(BC\). So \(AB > AC > BC\). Choice D incorrectly swaps \(BC\) and \(AC\), reversing the relationship between the \(30°\) and \(60°\) angles.

By the Triangle Midsegment Theorem, a segment connecting the midpoints of two sides of a triangle is parallel to the third side and exactly half its length. Since \(P\) and \(Q\) are midpoints of \(\overline{DE}\) and \(\overline{DF}\), we have \(PQ = \frac{1}{2}EF\) and \(\overline{PQ} \parallel \overline{EF}\). Choice A is wrong because \(\overline{PQ}\) is parallel (not perpendicular) to \(\overline{EF}\). Choice C incorrectly doubles the length instead of halving it. Choice D confuses a midsegment with an angle bisector.

The centroid lies \(\frac{2}{3}\) of the way from each vertex to the midpoint of the opposite side, giving a \(2:1\) ratio measured from the vertex. For example, if median \(\overline{AD}\) has length \(12\), then \(AG = 8\) and \(GD = 4\), yielding \(AG:GD = 2:1\). Choice A reverses the direction of measurement, which would describe a \(1:2\) ratio instead.

The Triangle Inequality Theorem requires the sum of any two sides to be strictly greater than the third. For \(\{6, 7, 12\}\): \(6 + 7 = 13 > 12\) ✓, \(6 + 12 = 18 > 7\) ✓, \(7 + 12 = 19 > 6\) ✓. For \(\{3, 4, 8\}\): \(3 + 4 = 7 < 8\), which fails. For \(\{5, 5, 10\}\): \(5 + 5 = 10\), which is not strictly greater than \(10\) (degenerate). For \(\{1, 3, 5\}\): \(1 + 3 = 4 < 5\), which fails.

The incenter is defined as the intersection of the three interior angle bisectors of a triangle. It is equidistant from all three sides and serves as the center of the inscribed circle. The three medians meet at the centroid (choice A). The three altitudes meet at the orthocenter (choice B). The three perpendicular bisectors of the sides meet at the circumcenter (choice D).

In any triangle, the longest side is opposite the largest angle. The largest angle is \(\angle Q = 70°\), and the side opposite \(\angle Q\) is \(\overline{PR}\). Therefore \(\overline{PR}\) is the longest side. By the same reasoning, \(\overline{QR}\) is opposite \(\angle P = 50°\) (the smallest angle), making it the shortest side, and \(\overline{PQ}\) is opposite \(\angle R = 60°\).

By definition, a median connects a vertex of a triangle to the midpoint of the opposite side. Every triangle has exactly three medians, and they all meet at the centroid. Choice A describes an altitude. Choice C describes the point determined by the Angle Bisector Theorem, not a median. Choice D describes part of a midsegment construction.

By the Triangle Midsegment Theorem, the midsegment connecting the midpoints of \(\overline{RS}\) and \(\overline{RT}\) is parallel to \(\overline{ST}\) and has length \(\frac{1}{2} \cdot ST\). Setting \(\frac{1}{2} \cdot ST = 9\) gives \(ST = 18\). Choice A (\(4.5\)) results from halving the midsegment length a second time. Choice B incorrectly treats the midsegment length and the side length as equal.

The centroid is the intersection point of the three medians of a triangle and always lies inside the triangle. It also serves as the triangle's center of mass. The angle bisectors meet at the incenter (choice A). The altitudes meet at the orthocenter (choice B). The perpendicular bisectors of the sides meet at the circumcenter (choice D).

The third side \(x\) must satisfy \(|15 - 8| < x < 15 + 8\), which gives the strict inequality \(7 < x < 23\). Only \(10\) satisfies this: \(7 < 10 < 23\). The value \(5\) fails because \(5 \leq 7\). The value \(7\) fails because the inequality requires \(x\) to be strictly greater than \(7\) (equality produces a degenerate segment, not a triangle). The value \(23\) fails because \(23\) is not strictly less than \(23\).

The centroid has coordinates \(\left(\frac{x_1 + x_2 + x_3}{3},\ \frac{y_1 + y_2 + y_3}{3}\right)\). Substituting: \(x = \frac{0 + 6 + 3}{3} = \frac{9}{3} = 3\) and \(y = \frac{0 + 0 + 9}{3} = \frac{9}{3} = 3\), giving centroid \((3, 3)\). Choice A gives only the midpoint of \(\overline{AB}\), not the centroid. Choices C and D result from averaging only two of the three vertices.

By the Angle Bisector Theorem, \(\frac{BD}{DC} = \frac{AB}{AC} = \frac{6}{9} = \frac{2}{3}\). Since \(BD + DC = BC = 10\), we write \(BD = \frac{2}{5} \cdot 10 = 4\) and \(DC = \frac{3}{5} \cdot 10 = 6\). A common error (choice C) is splitting \(BC\) in half, forgetting that the ratio depends on the adjacent sides \(AB\) and \(AC\), not on \(BC\) alone.

Each side of the midsegment triangle is a midsegment of triangle $ABC$, and by the Midsegment Theorem each midsegment equals half the corresponding side. Therefore the perimeter of the midsegment triangle equals \(\frac{1}{2}\) the perimeter of $ABC$. Setting $\frac{1}{2} \cdot P_{ABC} = 18$ gives $P_{ABC} = 36$. Choice A (\(9\)) halves the midsegment perimeter a second time. Choice B incorrectly treats the two perimeters as equal.

Apply the Triangle Inequality to each pair. The binding constraint is: \(x + (2x - 1) > x + 3\), which simplifies to \(3x - 1 > x + 3\), then \(2x > 4\), so \(x > 2\). The other two inequalities, \(x + (x+3) > 2x - 1\) and \((x+3) + (2x-1) > x\), reduce to \(3 > -1\) and \(2x > -2\) respectively, both of which hold for all positive \(x\). Also, \(2x - 1 > 0\) requires \(x > \frac{1}{2}\), which is weaker than \(x > 2\). Therefore the complete condition is \(x > 2\).

The centroid divides each median in a \(2:1\) ratio from vertex to midpoint. Therefore \(AG = \frac{2}{3} \cdot 15 = 10\) and \(GD = \frac{1}{3} \cdot 15 = 5\). Choice C (\(7.5\)) results from halving the median, which would be correct only if \(G\) were the midpoint of the median rather than a point \(\frac{1}{3}\) of the way from \(D\). Choice D (\(10\)) is the length \(AG\), not \(GD\).

By the Angle Bisector Theorem, \(\frac{QS}{SR} = \frac{PQ}{PR}\). Substituting known values: \(\frac{4}{6} = \frac{8}{PR}\). Cross-multiplying gives \(4 \cdot PR = 48\), so \(PR = 12\). Choice C (\(16\)) results from writing the ratio as \(\frac{QS}{SR} = \frac{PR}{PQ}\) (inverting the right-hand side). Choice D results from setting \(\frac{QS}{SR} = \frac{PQ}{PR}\) but solving incorrectly.

The three medians of any triangle are concurrent at the centroid, which always lies inside the triangle — this holds regardless of whether the triangle is acute, right, or obtuse. Choice A is false: medians are equal in length only in equilateral triangles. Choice B describes altitudes, not medians (a median targets a midpoint, not a perpendicular foot). Choice D describes angle bisectors.

In any triangle, larger sides are opposite larger angles. Side \(AB = 5\) is opposite \(\angle C\), side \(BC = 9\) is opposite \(\angle A\), and side \(AC = 7\) is opposite \(\angle B\). Since \(AB < AC < BC\) (that is, \(5 < 7 < 9\)), the opposite angles satisfy \(\angle C < \angle B < \angle A\). A common error is confusing which angle is opposite which side: the angle at vertex \(A\) is opposite side \(BC\), not any side touching \(A\).

Each side of the midsegment triangle equals \(\frac{1}{2}\) the corresponding side of triangle $ABC$, so the two triangles are similar with ratio \(\frac{1}{2}\). Since area scales as the square of the linear ratio, the midsegment triangle has area \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\) that of triangle $ABC$. Therefore $K = \frac{1}{4} \cdot [ABC]$, giving $[ABC] = 4K$. Choice A (\(2K\)) confuses the linear scale factor with the area scale factor.

The three medians of a triangle divide it into six smaller triangles of equal area. Each small triangle has area \(\frac{72}{6} = 12\) square units. Triangle $AGB$ (with vertices at \(A\), centroid \(G\), and \(B\)) consists of exactly two of these six equal triangles, giving an area of \(2 \times 12 = 24\) square units. Choice D (\(36\)) is the area of half the triangle, as each individual median divides the triangle into two halves of equal area — but triangle $AGB$ is not one such half.

By the Angle Bisector Theorem, \(\frac{AD}{DB} = \frac{CA}{CB}\). Substituting: \(\frac{4}{6} = \frac{10}{CB}\). Cross-multiplying gives \(4 \cdot CB = 60\), so \(CB = 15\). Choice C (\(20\)) results from inverting the left-hand ratio, writing \(\frac{DB}{AD} = \frac{CA}{CB}\) instead. Choice D (\(\frac{40}{3}\)) results from swapping the positions of \(CA\) and \(CB\) in the proportion.

We need \(\sqrt{15} - \sqrt{7} < k < \sqrt{7} + \sqrt{15}\). Since \(\sqrt{7} \approx 2.646\) and \(\sqrt{15} \approx 3.873\): the upper bound is \(\sqrt{7} + \sqrt{15} \approx 6.519\), so \(k \leq 6\); and the lower bound is \(\sqrt{15} - \sqrt{7} \approx 1.227\), so \(k \geq 2\). The valid integers are \(k \in \{2, 3, 4, 5, 6\}\), giving \(5\) values. Choice A (\(4\)) results from rounding the lower bound up to \(2\) but the upper bound down to \(5\), incorrectly excluding \(k = 6\).

The centroid divides each median in a \(2:1\) ratio from vertex to midpoint. For median \(\overline{BE}\) with total length \(18\): the longer segment is \(BG = \frac{2}{3} \cdot 18 = 12\) and the shorter segment is \(GE = \frac{1}{3} \cdot 18 = 6\). The information \(AG = 8\) is consistent (it confirms \(AD = 12\)) but is not needed to find \(GE\). A common error is computing \(GE = \frac{2}{3} \cdot 18 = 12\) by confusing the vertex-to-centroid segment (the longer part) with the centroid-to-midpoint segment (the shorter part).

Find the midpoints: \(M = \left(\frac{0+8}{2},\ \frac{0+0}{2}\right) = (4, 0)\) and \(N = \left(\frac{0+4}{2},\ \frac{0+6}{2}\right) = (2, 3)\). Then \(MN = \sqrt{(4-2)^2 + (0-3)^2} = \sqrt{4 + 9} = \sqrt{13}\). This can be verified by the Midsegment Theorem: \(ST = \sqrt{(8-4)^2 + (0-6)^2} = \sqrt{52} = 2\sqrt{13}\), and \(MN = \frac{1}{2} \cdot ST = \sqrt{13}\) ✓. Choice C (\(2\sqrt{13}\)) is the length of \(\overline{ST}\) itself, not the midsegment.

Use the incenter angle formula: $\angle BIC = 90° + \frac{\angle A}{2}$. With \(\angle A = 50°\): $\angle BIC = 90° + 25° = 115°$. This formula is derived from the fact that $\angle IBC = \frac{\angle B}{2}$ and $\angle ICB = \frac{\angle C}{2}$, so in triangle $BIC$: $\angle BIC = 180° - \frac{\angle B}{2} - \frac{\angle C}{2} = 180° - \frac{\angle B + \angle C}{2} = 180° - \frac{180° - 50°}{2} = 90° + 25° = 115°$. Choice B (\(105°\)) results from using \(90° + \angle A\) instead of \(90° + \frac{\angle A}{2}\).

The centroid formula gives \(G = \left(\frac{x_A + x_B + x_C}{3},\ \frac{y_A + y_B + y_C}{3}\right)\). For the \(x\)-coordinate: \(3 = \frac{0 + 6 + x_C}{3}\), so \(9 = 6 + x_C\) and \(x_C = 3\). For the \(y\)-coordinate: \(4 = \frac{0 + 3 + y_C}{3}\), so \(12 = 3 + y_C\) and \(y_C = 9\). Therefore \(C = (3, 9)\). Choice B (\(4, 9\)) results from a sign error when solving for \(x_C\). Choice D (\(6, 9\)) comes from incorrectly rearranging the centroid equation for the \(x\)-coordinate.

By the Triangle Midsegment Theorem, the segment connecting the midpoints of two sides is parallel to the third side and equals half its length. Since \(D\) and \(E\) are midpoints of \(\overline{AB}\) and \(\overline{AC}\), we get \(DE = \frac{1}{2}(BC) = \frac{1}{2}(18) = 9\). A common error is dividing by 3 instead of 2, giving 6, but the midsegment is always half of the parallel side.

By the Triangle Inequality Theorem, the sum of any two sides must be strictly greater than the third. For \(3, 7, 11\): \(3 + 7 = 10 < 11\), so these lengths cannot form a triangle. The other sets all pass: \(4+6=10>9\), \(5+5=10>9\), and \(6+8=14>12\).

The centroid divides each median in a \(2:1\) ratio, with the longer portion closer to the vertex. If a median has total length \(m\), then the vertex-to-centroid segment is \(\frac{2}{3}m\) and the centroid-to-midpoint segment is \(\frac{1}{3}m\). The ratio \(1:2\) is a common distractor because students sometimes reverse which segment is longer.

The Angle Bisector Theorem states that the bisector from a vertex divides the opposite side in the ratio of the two sides that form that angle. Since \(\overline{AD}\) bisects \(\angle A\), we have \(\frac{BD}{DC} = \frac{AB}{AC}\). Choice \(\frac{AB}{BC}\) is a common error that uses \(BC\) (the divided side itself) rather than the adjacent side \(AC\).

A median connects a vertex to the midpoint of the opposite side. Choice A describes an altitude, Choice C describes a midsegment, and Choice D describes an angle bisector. These four types of special segments are easily confused, so their precise definitions are worth memorizing.

A triangle has exactly 3 midsegments, one connecting each pair of side midpoints. By the Triangle Midsegment Theorem, each midsegment is parallel to the third (non-adjacent) side and equal to half its length. Midsegments do not pass through vertices — segments from vertices to opposite midpoints are medians, not midsegments.

The centroid divides each median in a \(2:1\) ratio from the vertex. So \(AG = \frac{2}{3}(15) = 10\) and \(GM = \frac{1}{3}(15) = 5\). Choice \(7.5\) is wrong because it treats the centroid as the midpoint of the median (a \(1:1\) ratio). The question asks for the shorter segment \(GM\), which is \(\frac{1}{3}\) of the total, not \(\frac{2}{3}\).

By the Midsegment Theorem, \(DE = \frac{1}{2}(BC)\), so \(3x - 1 = \frac{1}{2}(4x + 6) = 2x + 3\). Solving gives \(x = 4\). Verify: \(DE = 3(4)-1 = 11\) and \(BC = 4(4)+6 = 22 = 2(11)\). Choosing \(x = 3\) is incorrect; it yields \(DE = 8\) but \(BC = 18\), and \(8 \neq \frac{1}{2}(18) = 9\).

By the Triangle Inequality, the third side must satisfy \(|15 - 8| < s < 15 + 8\), giving \(7 < s < 23\). The inequalities must be strict because \(s = 7\) or \(s = 23\) would produce a degenerate triangle — three collinear points with zero area. Choice \(0 < s < 23\) ignores the lower bound imposed by the difference of the other two sides.

The centroid is the average of the three vertex coordinates: $G = \left(\frac{x_A + x_B + x_C}{3},\ \frac{y_A + y_B + y_C}{3}\right) = \left(\frac{1+7+4}{3},\ \frac{3+1+8}{3}\right) = \left(\frac{12}{3},\ \frac{12}{3}\right) = (4, 4)$. Choice \((3, 4)\) results from averaging only two of the three \(x\)-coordinates.

By the Angle Bisector Theorem, \(\frac{BD}{DC} = \frac{AB}{AC} = \frac{6}{10} = \frac{3}{5}\). Since \(BD + DC = 16\) with ratio \(3:5\), we get \(BD = \frac{3}{8}(16) = 6\). Choice \(8\) is the distractor from assuming the bisector hits the midpoint of \(\overline{BC}\) (which only occurs when \(AB = AC\)).

Each side of the midsegment triangle is a midsegment of the original triangle, so by the Midsegment Theorem each side equals half the length of the corresponding parallel side. Since all three sides scale by \(\frac{1}{2}\), the perimeter of the midsegment triangle is \(\frac{1}{2}(48) = 24\). Choice \(12\) incorrectly uses a factor of \(\frac{1}{4}\), which applies to area, not perimeter.

The centroid divides each median in a \(2:1\) ratio from the vertex, so \(BG = \frac{2}{3} \cdot BM = \frac{2}{3}(21) = 14\). Choice \(7 = \frac{1}{3}(21)\) gives the shorter segment \(GM\), not \(BG\). Choice \(10.5\) mistakenly uses \(\frac{1}{2}\) of the median length.

By the Angle Bisector Theorem, \(\frac{QS}{SR} = \frac{PQ}{PR} = \frac{8}{12} = \frac{2}{3}\). Setting up: \(\frac{x}{x+3} = \frac{2}{3}\), so \(3x = 2(x+3) = 2x + 6\), giving \(x = 6\). Therefore \(QS = 6\), \(SR = 9\), and \(QR = 15\). Choice \(12\) is the value of \(PR\), a plausible but incorrect answer.

Test all three triangle inequality conditions. \(x + 5 > x+2\) simplifies to \(3 > 0\) (always true), and \((x+2)+5 > x\) simplifies to \(7 > 0\) (always true). The binding condition is \(x + (x+2) > 5\), giving \(2x + 2 > 5\), so \(2x > 3\) and \(x > \frac{3}{2}\). Choice \(x > 1\) is too permissive: at \(x = 1.2\), for example, \(1.2 + 3.2 = 4.4 < 5\), which fails the inequality.

A key theorem states that in a right triangle, the median drawn to the hypotenuse equals half the hypotenuse. This holds because the midpoint of the hypotenuse is the circumcenter of the right triangle, equidistant from all three vertices. Therefore \(FM = \frac{1}{2}(DE) = \frac{1}{2}(26) = 13\). This result is independent of the leg lengths; knowing only the hypotenuse is sufficient.

Compute the midpoints: \(M = \left(\frac{0+10}{2}, \frac{0+0}{2}\right) = (5, 0)\) and \(N = \left(\frac{0+4}{2}, \frac{0+8}{2}\right) = (2, 4)\). Then \(MN = \sqrt{(5-2)^2 + (0-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\). This is confirmed by the Midsegment Theorem: \(BC = \sqrt{36 + 64} = 10\), so \(MN = \frac{1}{2}(10) = 5\). The distractor \(\sqrt{29}\) arises from computing \(\sqrt{(5-2)^2+(0-4)^2}\) with an arithmetic error.

A fundamental property of the centroid is that the three medians divide a triangle into exactly six smaller triangles of equal area. Therefore each smaller triangle has area \(\frac{30}{6} = 5\). Choice \(10\) arises from dividing by 3 (the number of medians, not the number of smaller triangles). Choice \(6\) has no geometric justification here.

First find \(\angle A\) using the law of cosines: \(\cos A = \frac{AC^2 + AB^2 - BC^2}{2 \cdot AC \cdot AB} = \frac{64 + 25 - 49}{80} = \frac{40}{80} = \frac{1}{2}\), so \(A = 60^\circ\) and \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\). Using the angle bisector length formula \(t_a = \frac{2bc\cos(A/2)}{b+c}\) with \(b = AC = 8\) and \(c = AB = 5\): \(AD = \frac{2(8)(5) \cdot \frac{\sqrt{3}}{2}}{13} = \frac{40\sqrt{3}}{13}\). Choice \(\frac{40}{13}\) is a distractor that drops the \(\sqrt{3}\) factor.

Using the median length formula \(m_b^2 = \frac{2a^2 + 2c^2 - b^2}{4}\), where \(a = BC = 5\), \(b = CA = 8\), and \(c = AB = 7\): \(m_b^2 = \frac{2(25) + 2(49) - 64}{4} = \frac{50 + 98 - 64}{4} = \frac{84}{4} = 21\), so \(m_b = \sqrt{21}\). A common error leading to \(\sqrt{29}\) is omitting the \(-b^2\) term and computing \(\frac{50+98}{4} = \frac{148}{4}\), which does not simplify correctly.

By the Midsegment Theorem, each side of the midsegment triangle is half the length of the corresponding side of triangle $ABC$. When all linear dimensions scale by \(\frac{1}{2}\), area scales by \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\). Therefore the original triangle has area \(4 \times 16 = 64\). Choice \(32\) is incorrect because it uses a linear scale factor of \(2\) instead of the quadratic factor of \(4\) needed for area.

The incenter is at \(\frac{a \cdot A + b \cdot B + c \cdot C}{a+b+c}\), where \(a, b, c\) are the side lengths opposite to \(A, B, C\). Here \(a = BC = \sqrt{36+64} = 10\), \(b = CA = 8\), \(c = AB = 6\). Incenter \(= \frac{10(0,0)+8(6,0)+6(0,8)}{24} = \frac{(48,48)}{24} = (2,2)\). Alternatively, for a right triangle with legs \(6\) and \(8\) and hypotenuse \(10\), the inradius is \(r = \frac{6+8-10}{2} = 2\), and since both legs lie on the axes, the incenter is at \((r, r) = (2, 2)\).

Since the centroid divides each median in a \(2:1\) ratio from the vertex, \(AG = 2 \cdot GM\). Setting up the equation: \(3x - 2 = 2(x + 1) = 2x + 2\), which gives \(x = 4\). Then \(AG = 3(4) - 2 = 10\) and \(GM = 4 + 1 = 5\), so \(AM = 10 + 5 = 15\). Choice \(12\) results from incorrectly assuming \(AG = GM\) (a \(1:1\) ratio), which would give \(3x-2 = x+1\), so \(x = 1.5\) and \(AM = 5\).

By the Angle Bisector Theorem, \(\frac{AD}{DB} = \frac{CA}{CB}\), so \(\frac{3}{5} = \frac{CA}{CB}\). Let \(CA = 3k\) and \(CB = 5k\). Using the perimeter: \(AB + CA + CB = 40 \Rightarrow 8 + 3k + 5k = 40 \Rightarrow 8k = 32 \Rightarrow k = 4\). Therefore \(CB = 5(4) = 20\) and \(CA = 3(4) = 12\). Choice \(16\) is a numerical coincidence equal to \(2 \cdot AB\) and has no geometric basis here.

By the Triangle Inequality, \(n\) must satisfy \(|18 - 12| < n < 18 + 12\), giving \(6 < n < 30\). The positive integers in this open interval are \(7, 8, 9, \ldots, 29\), for a count of \(29 - 7 + 1 = 23\). The boundary values \(n = 6\) and \(n = 30\) are excluded because they produce degenerate (zero-area) figures. Choice \(24\) incorrectly includes one boundary value.