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Geometry

Geometry Help: Theorems About Triangles

Review real example questions for Theorems About Triangles in Geometry.

Question 1

In triangle $ABC$ , point $D$ lies on side $\overline{BC}$ , and $\overline{AD}$ is the perpendicular bisector of side $\overline{BC}$ . If $AB = 13$ and $BD = 5$ , what is the length of side $\overline{AC}$ ?

$12$
$13$
$10$
$8$

Explanation: 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.

Question 2

In the diagram, $\triangle ABC$ is shown in the plane. Segments $AB$ and $AC$ have matching single tick marks, indicating they are congruent. No angle arcs, parallel marks, right-angle boxes, midpoint markings, or lengths are given, and the diagram is not drawn to scale. Which statement must be true?

$\angle ABC \cong \angle ACB$
$BC$ is perpendicular to $AB$
$B$ is the midpoint of $AC$
$AB \parallel AC$

Explanation: 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.

Question 3

In the diagram, $\triangle PQR$ is shown. Point $M$ lies on segment $PQ$ and point $N$ lies on segment $PR$ . The markings show $PM \cong MQ$ (matching tick marks on $PM$ and $MQ$ ) and $PN \cong NR$ (matching tick marks on $PN$ and $NR$ ). Segment $MN$ is drawn. No parallel arrows, angle markings, or lengths are given, and the diagram is not drawn to scale. Which conclusion follows from the diagram?

$MN \parallel QR$
$MN \perp QR$
$M$ is the midpoint of $QR$
$\angle PMN \cong \angle PNM$

Explanation: 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.

Question 4

Triangle $PQR$ is shown in the plane. Point $M$ lies on segment $PQ$ and point $N$ lies on segment $PR$ . The diagram marks $PM \cong MQ$ (matching tick marks on the two parts of $PQ$ ) and $PN \cong NR$ (matching tick marks on the two parts of $PR$ ). Segment $MN$ is drawn. No angle measures, no parallel markings, and no lengths are given, and the diagram is not drawn to scale.

Which statement must be true?

$MN \parallel QR$
$MN \perp QR$
$M$ is the midpoint of $QR$
$MN \cong QR$

Explanation: 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.

Question 5

In the diagram, $\triangle RST$ is shown. Point $M$ lies on $\overline{ST}$ with midpoint markings indicating $SM\cong MT$ . Segment $\overline{RM}$ is drawn. No other markings are given, and the diagram is not drawn to scale.

Which relationship can be proven from the diagram?

$\overline{RM}$ is a midsegment of $\triangle RST$
$\overline{RM}$ is a median of $\triangle RST$
$\overline{RM}\parallel \overline{ST}$
$\angle RSM\cong \angle MRT$

Explanation: 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.

Question 6

In the diagram, triangle $ABC$ is shown in the plane. Segment $AB$ and segment $AC$ have matching single tick marks, indicating they are congruent. No angle measures, parallel markings, right-angle markings, or midpoint markings are shown, and the diagram is not drawn to scale.

Which conclusion follows from the diagram?

$\angle ABC \cong \angle ACB$
$BC \cong AB$
$AD \perp BC$
$BD \cong DC$

Explanation: 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.

Question 7

Point $R$ is equidistant from points $S$ and $T$ . Point $R$ lies on line $\ell$ , and line $\ell$ is perpendicular to segment $\overline{ST}$ at point $U$ . If $SU = 3x - 4$ and $UT = 2x + 6$ , what is the value of $x$ ?

$10$
$2$
$5$
$8$

Explanation: 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

Question 8

In the diagram, $\triangle JKL$ is shown. Segment $JM$ is drawn from vertex $J$ to point $M$ on $KL$ , and segment $KN$ is drawn from vertex $K$ to point $N$ on $JL$ . The two segments intersect at point $X$ . Markings indicate $KM \cong ML$ and $JN \cong NL$ . No other markings are shown (no right-angle box, no angle arcs, no parallel arrows, no lengths), and the diagram is not drawn to scale. Which relationship can be proven?

$X$ is the centroid of $\triangle JKL$
$JM \perp KL$
$\angle JKL \cong \angle JLK$
$X$ is the incenter of $\triangle JKL$

Explanation: 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.

Question 9

In the diagram, $\triangle XYZ$ is shown. Segments $XW$ and $YV$ are drawn from vertices $X$ and $Y$ to points $W$ on $YZ$ and $V$ on $XZ$ , respectively. Markings indicate $YW \cong WZ$ and $XV \cong VZ$ . The segments intersect at point $G$ . No other markings are shown (no angle arcs, no right-angle boxes, no parallel arrows, no lengths), and the diagram is not drawn to scale. Which statement must be true?

$G$ is the centroid of $\triangle XYZ$
$G$ is the circumcenter of $\triangle XYZ$
$XW$ is perpendicular to $YZ$
$\angle XYZ \cong \angle XZY$

Explanation: 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.

Question 10

In $\triangle PQR$ (shown), segments $\overline{PQ}$ and $\overline{PR}$ have matching tick marks indicating $PQ\cong PR$ .

Which statement must be true?

$\angle PRQ\cong \angle PQR$
$\angle QPR$ is a right angle
$QR\cong PQ$
$\overline{QR}$ bisects $\angle QPR$

Explanation: 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.

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Geometry

Geometry Help: Theorems About Triangles

Review real example questions for Theorems About Triangles in Geometry.

Question 1

$12$
$13$
$10$
$8$

Explanation: 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.

Question 2

$\angle ABC \cong \angle ACB$
$BC$ is perpendicular to $AB$
$B$ is the midpoint of $AC$
$AB \parallel AC$

Question 3

$MN \parallel QR$
$MN \perp QR$
$M$ is the midpoint of $QR$
$\angle PMN \cong \angle PNM$

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

Question 4

Which statement must be true?

$MN \parallel QR$
$MN \perp QR$
$M$ is the midpoint of $QR$
$MN \cong QR$

Question 5

Which relationship can be proven from the diagram?

$\overline{RM}$ is a midsegment of $\triangle RST$
$\overline{RM}$ is a median of $\triangle RST$
$\overline{RM}\parallel \overline{ST}$
$\angle RSM\cong \angle MRT$

Question 6

Which conclusion follows from the diagram?

$\angle ABC \cong \angle ACB$
$BC \cong AB$
$AD \perp BC$
$BD \cong DC$

Question 7

$10$
$2$
$5$
$8$

Explanation: 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

Question 8

$X$ is the centroid of $\triangle JKL$
$JM \perp KL$
$\angle JKL \cong \angle JLK$
$X$ is the incenter of $\triangle JKL$

Question 9

$G$ is the centroid of $\triangle XYZ$
$G$ is the circumcenter of $\triangle XYZ$
$XW$ is perpendicular to $YZ$
$\angle XYZ \cong \angle XZY$

Question 10

In $\triangle PQR$ (shown), segments $\overline{PQ}$ and $\overline{PR}$ have matching tick marks indicating $PQ\cong PR$ .

Which statement must be true?

$\angle PRQ\cong \angle PQR$
$\angle QPR$ is a right angle
$QR\cong PQ$
$\overline{QR}$ bisects $\angle QPR$