Geometry Quiz: Theorems About Triangles

20 questions · exam conditions

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Theorems About TrianglesQuestion 1 of 20

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

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Geometry Quiz

Geometry Quiz: Theorems About Triangles

Practice Theorems About Triangles in Geometry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Theorems About Triangles, giving you a quick way to practice the rules, question types, and explanations that matter most for Geometry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$12$
$13$ (correct answer)
$10$
$8$

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?

$\angle ABC \cong \angle ACB$ (correct answer)
$BC$ 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?

$MN \parallel QR$ (correct answer)
$MN \perp QR$
$M$ 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?

$MN \parallel QR$ (correct answer)
$MN \perp QR$
$M$ 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 $\triangle RST$ (correct answer)
$\overline{RM}\parallel \overline{ST}$
$\angle RSM\cong \angle MRT$

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?

$\angle ABC \cong \angle ACB$ (correct answer)
$BC \cong AB$
$AD \perp BC$
$BD \cong DC$

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 $x$ ?

$10$ (correct answer)
$2$
$5$
$8$

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 $\triangle JKL$ (correct answer)
$JM \perp KL$
$\angle JKL \cong \angle JLK$
$X$ 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 $\triangle XYZ$ (correct answer)
$G$ 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?

$\angle PRQ\cong \angle PQR$ (correct answer)
$\angle QPR$ 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.

Question 11

In the diagram, $\triangle RST$ is shown. Point $U$ lies on $RS$ with markings indicating $RU \cong US$ , and point $V$ lies on $ST$ with markings indicating $SV \cong VT$ . Segment $UV$ is drawn. No angle markings, no right-angle boxes, no parallel arrows, and no lengths are given; the diagram is not drawn to scale. Which statement must be true?

$UV \parallel RT$ (correct answer)
$UV \perp RT$
$U$ is the midpoint of $RT$
$\angle URV \cong \angle VTR$

Explanation: This question involves theorems about triangles, specifically the midsegment theorem. Conceptually, it states that joining midpoints of two sides creates a segment parallel to the third. Markings indicate RU ≅ US and SV ≅ VT, showing U and V as midpoints on RS and ST. In triangle RST, UV applies as the midsegment parallel to RT. The conclusion is justified by the theorem's parallelism property. A misconception in distractors is assuming midpoints imply angle congruence without further evidence. To transfer this strategy, match midpoint markings to known triangle theorems like the midsegment for parallelism.

Question 12

Point $P$ lies on the perpendicular bisector of segment $\overline{MN}$ . If $PM = 2x + 7$ and $PN = 3x - 1$ , and the perpendicular bisector intersects $\overline{MN}$ at point $Q$ where $MQ = x + 5$ , what is the length of $\overline{MN}$ ?

$26$ (correct answer)
$13$
$21$
$23$

Explanation: Since

P

is on the perpendicular bisector of

\overline{MN}

, by the perpendicular bisector theorem,

PM = PN

. So

2x + 7 = 3x - 1

, giving us

8 = x

. Since

Q

is the midpoint of

\overline{MN}

, we have

MQ = QN = x + 5 = 8 + 5 = 13

. Therefore,

MN = 2 \cdot 13 = 26

. Choice B (13) is just the length of

MQ

, not the full segment. Choice C (21) comes from

PM = 2(8) + 7 = 23

minus 2. Choice D (23) is the value of

PM

when

x = 8

Question 13

In triangle $DEF$ , the perpendicular bisector of side $\overline{DE}$ passes through vertex $F$ . If $DE = 16$ and the perpendicular bisector intersects $\overline{DE}$ at point $M$ , which of the following must be true?

Triangle $DEF$ is equilateral with all sides equal to 16
Triangle $DEF$ is scalene with $FM$ being the longest side
Triangle $DEF$ is a right triangle with the right angle at $M$
Triangle $DEF$ is isosceles with $FD = FE$ , and $DM = ME = 8$ (correct answer)

Explanation: When you encounter a problem involving perpendicular bisectors passing through vertices, you're dealing with a fundamental property of isosceles triangles. The key insight is that any point on a perpendicular bisector is equidistant from the endpoints of the bisected segment. Since the perpendicular bisector of

\overline{DE}

passes through vertex

F

, point

F

lies on this perpendicular bisector. By the perpendicular bisector theorem, any point on the perpendicular bisector of a segment is equidistant from the endpoints of that segment. Therefore,

FD = FE

, making triangle

DEF

isosceles. Since

M

is where the perpendicular bisector intersects

\overline{DE}

, point

M

bisects

\overline{DE}

, so

DM = ME = \frac{16}{2} = 8

. Choice A is incorrect because knowing that

FD = FE

doesn't tell us anything about the length of side

\overline{DE}

relative to the other sides. The triangle could be isosceles but not equilateral. Choice B is wrong because we've established the triangle is isosceles (

FD = FE

), not scalene. Also,

FM

is a height of the triangle, not a side. Choice C misunderstands the geometry. While the perpendicular bisector creates a right angle with

\overline{DE}

at point

M

, this doesn't make triangle

DEF

a right triangle. Point

M

is not even a vertex of triangle

DEF

. Remember: whenever a perpendicular bisector of one side of a triangle passes through the opposite vertex, that triangle must be isosceles with the two sides adjacent to the bisected side being equal.

Question 14

In the diagram, $\triangle DEF$ is shown. Point $G$ lies on $EF$ with markings indicating $EG \cong GF$ , and point $H$ lies on $DF$ with markings indicating $DH \cong HF$ . Segment $GH$ is drawn. No angle markings, no right-angle boxes, no parallel arrows, and no lengths are given; the diagram is not drawn to scale. Which statement correctly describes the segment shown?

$GH \parallel ED$ (correct answer)
$GH$ bisects $\angle DFE$
$GH$ is an altitude from $G$ to $DF$
$GH \cong ED$

Explanation: This question involves theorems about triangles, specifically the midsegment theorem. The midsegment theorem conceptually asserts that connecting midpoints of two sides forms a segment parallel to the third side. The marked features are matching ticks showing EG ≅ GF and DH ≅ HF, indicating G and H as midpoints of EF and DF. In triangle DEF, applying the theorem to midpoints G and H means GH is parallel to the third side ED. The justification lies in the proportional division of sides creating parallel lines via similarity. A misconception in distractors is assuming angle bisection or altitude without angle or perpendicular markings. To transfer this strategy, match midpoint markings to known triangle theorems such as midsegment for identifying parallelism.

Question 15

Two intersecting lines form four angles at their point of intersection. One pair of vertical angles has measures $(3n + 15)°$ and $(5n - 25)°$ . What is the measure of each angle in the other pair of vertical angles?

$75°$
$120°$
$105°$ (correct answer)
$60°$

Explanation: When two lines intersect, they create four angles at their intersection point. The key principle here is that vertical angles (opposite angles) are always equal, while adjacent angles (next to each other) are supplementary and add up to

180°

. Since vertical angles are equal, you can set up an equation:

(3n + 15)° = (5n - 25)°

. Solving for

n

3n + 15 = 5n - 25

15 + 25 = 5n - 3n

40 = 2n

n = 20

Substituting back:

(3(20) + 15)° = 75°

and

(5(20) - 25)° = 75°

. This confirms our calculation is correct. Now, since adjacent angles are supplementary, the other pair of vertical angles measures

180° - 75° = 105°

. Looking at the wrong answers: Choice A (

75°

) is the measure of the given pair of vertical angles, not the other pair. Choice B (

120°

) might result from incorrectly adding the two expressions instead of setting them equal. Choice D (

60°

) could come from calculation errors or misunderstanding the relationship between vertical angles. Study tip: Remember that intersecting lines create two pairs of vertical angles. Once you find one pair's measure, subtract from

180°

to find the other pair. Always verify that vertical angles are equal and adjacent angles sum to

180°

Question 16

In the diagram, $\triangle ABC$ is shown. Point $D$ is marked as the midpoint of $\overline{BC}$ (with $BD\cong DC$ ). Point $E$ is marked as the midpoint of $\overline{AC}$ (with $AE\cong EC$ ). Segment $\overline{DE}$ is drawn. No other markings are given, and the diagram is not drawn to scale.

Which statement must be true?

$\overline{DE}\parallel \overline{AB}$ (correct answer)
$\overline{DE}\parallel \overline{BC}$
$\angle ADE\cong \angle DEA$
$AB\cong AC$

Explanation: Theorems about triangles feature the midsegment theorem for midpoint parallels. Conceptually, it states the join of two sides' midpoints parallels the third side. Markings show D as midpoint of BC and E of AC. Applying the theorem, DE parallels AB as the third side. Justification is the theorem's application to mids from C. Distractor misconceptions assume parallelism to other sides, as in choice B. Transfer by matching midpoint pairs to midsegment for parallelism.

Question 17

In the diagram, $\triangle LMN$ is shown. Segments $\overline{LM}$ and $\overline{LN}$ have matching tick marks indicating $LM\cong LN$ . No other markings are given, and the diagram is not drawn to scale.

Which conclusion follows from the diagram?

$\angle LNM\cong \angle LMN$ (correct answer)
$MN\cong LM$
$\overline{MN}$ bisects $\angle L$
$\overline{LM}\perp \overline{LN}$

Explanation: Theorems about triangles include isosceles properties linking sides to angles. The isosceles theorem states congruent sides imply congruent opposite angles. Tick marks mark LM congruent to LN in the diagram. Applying it, base angles at M and N are congruent. Justification is the theorem's symmetry from vertex L. Distractor misconceptions assume side equality without basis, as in choice B. Transfer strategy: match side ticks to isosceles for angle congruences.

Question 18

In the diagram, $\triangle ABC$ is shown. Point $D$ lies on $\overline{BC}$ with midpoint markings indicating $BD\cong DC$ . Segment $\overline{AD}$ is drawn. No other markings are given, and the diagram is not drawn to scale.

Which statement must be true?

$\overline{AD}$ is a median of $\triangle ABC$ (correct answer)
$\overline{AD}\perp \overline{BC}$
$\angle BAD\cong \angle DAC$
$AB\cong AC$

Explanation: Theorems about triangles define medians as segments to midpoints of opposite sides. Conceptually, a median connects a vertex to the midpoint of the opposite side. The diagram's midpoint markings place D at the midpoint of BC. Applying this, AD is a median from A to BC's midpoint. Justification stems from the definition matching the configuration. Distractor misconceptions include assuming perpendicularity without markings, as in choice B. Transfer strategy: match midpoint to opposite vertex for median theorems.

Question 19

In the diagram, $\triangle ABC$ is shown. Segment $\overline{AB}$ and segment $\overline{AC}$ have matching tick marks indicating $AB\cong AC$ .

Which property is guaranteed by the markings?

$\angle BAC\cong \angle ACB$
$\angle ABC\cong \angle BCA$ (correct answer)
$BC\cong AB$
$\overline{BC}$ is perpendicular to $\overline{AB}$

Explanation: Theorems about triangles cover isosceles triangles, where side equality implies angle equality. The isosceles theorem conceptually notes that congruent sides opposite congruent base angles. Marked tick marks show AB congruent to AC in the diagram. The theorem applies, making base angles at B and C congruent. Justification arises from the symmetry of equal sides from vertex A. Distractor misconceptions include assuming vertex angle equality, as in choice A. Transfer strategy: match side congruences to isosceles theorems for base angles.

Question 20

In the diagram, $\triangle DEF$ is shown. Point $M$ lies on $\overline{DE}$ and point $N$ lies on $\overline{DF}$ . Midpoint markings indicate $DM\cong ME$ and $DN\cong NF$ .

Which statement must be true?

$\overline{MN}$ is an altitude to $\overline{EF}$
$\overline{MN}$ is an angle bisector of $\angle EDF$
$MN=\tfrac12\,EF$ (correct answer)
$\angle EMN\cong \angle ENM$

Explanation: Theorems about triangles involve the midsegment theorem for midpoint segments. This theorem states that a segment between midpoints of two sides is half the third side's length and parallel. Midpoint markings identify M on DE and N on DF as midpoints. Applying it, MN is the midsegment, equaling half of EF. Justification is the theorem's length proportion in midpoint connections. Distractor misconceptions assume altitudes without perpendiculars, as in choice A. Transfer by matching midpoints to midsegment theorem for lengths.