Since $$\overline{AD}$$ is the perpendicular bisector of $$\overline{BC}$$, by the perpendicular bisector theorem, any point on the perpendicular bisector is equidistant from the endpoints of the segment. Therefore, $$AB = AC = 13$$. Choice A (12) might result from incorrectly using the Pythagorean theorem with $$BD = 5$$ and assuming $$AD = 12$$. Choice C (10) could come from subtracting $$BD$$ from $$AB$$. Choice D (8) might result from misapplying distance relationships.

This question involves theorems about triangles, focusing on properties of isosceles triangles. The isosceles triangle theorem states that if two sides of a triangle are congruent, then the angles opposite those sides are also congruent. The diagram features matching tick marks on segments AB and AC, indicating they are congruent. Applying the theorem, since AB ≅ AC, the angles opposite them, which are angle ABC and angle ACB, must be congruent. This conclusion is justified because the equal sides create symmetry in the triangle, making the base angles equal. A common distractor misconception is assuming perpendicularity or midpoints without supporting markings, such as confusing side congruence with right angles. To transfer this strategy, always match markings like tick marks to known triangle theorems such as the isosceles base angles theorem.

This question involves theorems about triangles, particularly the midsegment theorem. The midsegment theorem states that the segment joining the midpoints of two sides of a triangle is parallel to the third side. The diagram shows matching tick marks indicating $PM \cong MQ$ and $PN \cong NR$, meaning M and N are midpoints of $PQ$ and $PR$ respectively. Applying the theorem in triangle $PQR$, segment $MN$ connects these midpoints and thus must be parallel to $QR$. This is justified because the midsegment creates a smaller triangle similar to the original, enforcing parallelism. A distractor misconception is assuming perpendicularity or angle congruence without evidence from markings. To transfer this strategy, match midpoint markings to known triangle theorems like the midsegment theorem for parallelism.

The skill involves theorems about triangles, focusing on properties of segments connecting midpoints. The midsegment theorem states that a segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length. The diagram identifies points M and N as midpoints of PQ and PR, respectively, with matching tick marks confirming PM congruent to MQ and PN congruent to NR. Applying the theorem, segment MN connects these midpoints, so it must be parallel to the third side QR. This conclusion is justified as the midsegment theorem directly applies to midpoints on two sides, ensuring parallelism. A distractor misconception might involve assuming perpendicularity without any right-angle indicators. To approach similar diagrams, match midpoint markings to theorems like the midsegment theorem for parallelism or length relationships.

Theorems about triangles distinguish medians from midsegments in midpoint usage. A median is conceptually a line from a vertex to the midpoint of the opposite side. Markings indicate M as the midpoint of ST in the diagram. Applying the definition, RM is a median from R to ST's midpoint. This is justified by the direct vertex-to-midpoint connection. Distractor misconceptions confuse medians with midsegments, as in choice A. Transfer by matching vertex-to-midpoint to median theorems.

The skill involves theorems about triangles, particularly those connecting side lengths to angle measures. The isosceles triangle theorem states that if two sides of a triangle are congruent, then the base angles opposite those sides are also congruent. The diagram features matching tick marks on segments AB and AC, indicating their congruence. Applying the theorem to triangle ABC, since AB is congruent to AC, the angles opposite them—angle ABC opposite AC and angle ACB opposite AB—must be congruent. This conclusion is justified because the theorem guarantees equal base angles in an isosceles triangle with AB and AC as the equal sides. A common distractor misconception is assuming a perpendicular bisector like AD without any right-angle or midpoint markings shown. To solve similar problems, match the diagram markings to known triangle theorems such as isosceles properties or congruence criteria.

Since $$R$$ is equidistant from $$S$$ and $$T$$, and $$R$$ lies on line $$\ell$$ which is perpendicular to $$\overline{ST}$$, line $$\ell$$ must be the perpendicular bisector of $$\overline{ST}$$. Therefore, $$U$$ bisects $$\overline{ST}$$, so $$SU = UT$$. Setting up the equation: $$3x - 4 = 2x + 6$$. Solving: $$3x - 2x = 6 + 4$$, so $$x = 10$$. Choice B (2) comes from solving $$3x - 4 = 2x + 6$$ incorrectly as $$x - 4 = 6$$. Choice C (5) might result from $$\frac{6+4}{2}$$. Choice D (8) could come from $$6 + 4 - 2$$.

This question involves theorems about triangles, centering on concurrency points like the centroid. The centroid theorem states that medians of a triangle intersect at a single point called the centroid, dividing each median in a 2:1 ratio. The diagram marks KM ≅ ML and JN ≅ NL, showing M and N as midpoints of KL and JL. Applying this, JM and KN are medians from J and K, intersecting at X, which must be the centroid. Justification comes from the property that all medians concur at the centroid, even if only two are shown. A distractor misconception is confusing the centroid with the incenter, which requires angle bisectors. To transfer this strategy, match midpoint markings on sides to known triangle theorems involving medians and centroids.

This question involves theorems about triangles, particularly those concerning the centroid. The centroid is conceptually the intersection point of the medians in a triangle. Markings show YW ≅ WZ and XV ≅ VZ, indicating W and V as midpoints of YZ and XZ. In triangle XYZ, XW and YV are medians intersecting at G, identifying G as the centroid. Justification stems from the concurrency of medians at the centroid. A distractor misconception is mistaking it for the circumcenter, which involves perpendicular bisectors. To transfer this strategy, match midpoint markings to known triangle theorems involving medians and centroids.

Theorems about triangles encompass isosceles triangle properties, where equal sides lead to equal base angles. The isosceles triangle theorem states that if two sides of a triangle are congruent, then the angles opposite those sides are congruent. In this diagram, the matching tick marks indicate that PQ is congruent to PR. Applying the theorem, the base angles at Q and R are congruent, so angle PRQ equals angle PQR. This is justified because the equal sides from vertex P create symmetry in the base angles. A distractor misconception is assuming a right angle without perpendicular markings, as in choice B. To transfer this, match congruent side markings to the isosceles triangle theorem for angle conclusions.