Math 2 Quiz: Writing Geometric Proofs

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Writing Geometric ProofsQuestion 1 of 19

In triangle $ABC$ , $AB = AC$ and $D$ is a point on $\overline{BC}$ such that $\overline{AD} \perp \overline{BC}$ . To prove that $\overline{AD}$ bisects $\overline{BC}$ , which statement correctly identifies the critical step in the proof?

Prove that triangles

ABD

and

ACD

are congruent using SAS, since

AB = AC

\angle ADB = \angle ADC = 90°

, and

AD = AD

Apply the isosceles triangle theorem to conclude that

\angle ABC = \angle ACB

, then use the fact that equal angles subtend equal sides

Use the Pythagorean theorem on triangles

ABD

and

ACD

to show that

BD^2 + AD^2 = CD^2 + AD^2

, therefore

BD = CD

Apply the perpendicular bisector theorem directly, since

\overline{AD}

is perpendicular to

\overline{BC}

and

A

is equidistant from

B

and

C

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Math 2 Quiz

Math 2 Quiz: Writing Geometric Proofs

Practice Writing Geometric Proofs in Math 2 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 Writing Geometric Proofs, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

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

Prove that triangles $ABD$ and $ACD$ are congruent using SAS, since $AB = AC$ , $\angle ADB = \angle ADC = 90°$ , and $AD = AD$ (correct answer)
Apply the isosceles triangle theorem to conclude that $\angle ABC = \angle ACB$ , then use the fact that equal angles subtend equal sides
Use the Pythagorean theorem on triangles $ABD$ and $ACD$ to show that $BD^2 + AD^2 = CD^2 + AD^2$ , therefore $BD = CD$
Apply the perpendicular bisector theorem directly, since $\overline{AD}$ is perpendicular to $\overline{BC}$ and $A$ is equidistant from $B$ and $C$

Explanation: The correct approach is to prove triangles ABD and ACD are congruent using SAS: AB = AC (given), ∠ADB = ∠ADC = 90° (given that AD ⊥ BC), and AD = AD (reflexive property). From congruence, BD = CD by CPCTC. Choice B doesn't directly lead to bisection. Choice C incorrectly applies the Pythagorean theorem. Choice D reverses the logic - we need to prove D bisects BC first before we can apply the perpendicular bisector theorem.

Question 2

In quadrilateral $WXYZ$ , the diagonals $\overline{WY}$ and $\overline{XZ}$ bisect each other at point $M$ . A student claims this is sufficient to prove that $WXYZ$ is a parallelogram. What additional information would make this reasoning invalid?

If the diagonals are not congruent, then the quadrilateral could be a rhombus instead of a general parallelogram
If the diagonals are perpendicular, then the quadrilateral must be a rhombus, which is a special case requiring different proof methods
No additional information would invalidate the reasoning, since diagonals bisecting each other is sufficient to prove a parallelogram (correct answer)
If the sides are not all congruent, then the quadrilateral cannot be proven to be a parallelogram using only diagonal bisection

Explanation: The student's reasoning is completely valid. One of the standard theorems states that if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. No additional conditions are needed. Choice A incorrectly suggests this affects the validity (a rhombus is still a parallelogram). Choice B incorrectly suggests perpendicular diagonals invalidate the proof (a rhombus is a special parallelogram). Choice D incorrectly states that equal sides are required for the theorem.

Question 3

A student is proving that if two parallel lines are cut by a transversal, then alternate interior angles are congruent. The student has established that lines $m$ and $n$ are parallel, and line $t$ is a transversal. What assumption would make this proof circular reasoning?

Assuming that corresponding angles formed by parallel lines and a transversal are congruent, then using linear pairs to prove alternate interior angles are congruent
Assuming that vertical angles are congruent, then using this fact along with the parallel postulate to establish the alternate interior angle relationship
Assuming that alternate interior angles are congruent when formed by parallel lines, then using this to prove the statement being demonstrated (correct answer)
Assuming that the sum of interior angles on the same side of the transversal is $180°$ , then using supplementary angle relationships

Explanation: Circular reasoning occurs when you assume what you're trying to prove. Choice C directly assumes alternate interior angles are congruent for parallel lines, which is exactly what the student is trying to prove. Choice A uses the related but distinct corresponding angles theorem. Choice B uses vertical angles (a valid given) and the parallel postulate. Choice D uses co-interior angles, which is a different theorem that can be used to prove alternate interior angles.

Question 4

In a proof that the base angles of an isosceles triangle are congruent, a student draws an altitude from the vertex angle to the base. What property of this altitude is crucial for completing the proof?

The altitude creates two right triangles that are congruent by SAS, since they share the altitude, have congruent hypotenuses from the equal legs, and form right angles at the base
The altitude bisects the vertex angle, creating two triangles that are congruent by ASA using the bisected angle, the altitude, and the right angles formed
The altitude bisects the base, creating two triangles that are congruent by SAS using the equal legs, the shared altitude, and the congruent base segments (correct answer)
The altitude creates similar right triangles with proportional corresponding sides, and angle-angle similarity guarantees that the acute angles are congruent

Explanation: In an isosceles triangle, the altitude from the vertex angle to the base bisects the base. This creates two right triangles that are congruent by SAS: the two legs of the isosceles triangle are congruent (given), the altitude is shared, and the two segments of the base are congruent (because altitude bisects the base). By CPCTC, the base angles are congruent. Choice A has the right congruence method but wrong reasoning. Choice B incorrectly identifies angle bisection as the crucial property. Choice D applies similarity when congruence is needed.

Question 5

Given rhombus $EFGH$ with diagonals $\overline{EG}$ and $\overline{FH}$ intersecting at point $J$ , a proof aims to show that $\overline{EG} \perp \overline{FH}$ . Which statement represents a logical gap that would weaken this proof?

Failing to establish that all sides of the rhombus are congruent before applying properties of rhombi to conclude perpendicular diagonals
Assuming that diagonals of any parallelogram are perpendicular, rather than specifically proving this property holds for rhombi (correct answer)
Not proving that the diagonals bisect each other before concluding that they are perpendicular, since bisection is required for perpendicularity
Failing to use coordinate geometry to calculate slopes and verify that the product of diagonal slopes equals $-1$ for perpendicularity

Explanation: Choice B identifies the critical error: assuming that all parallelograms have perpendicular diagonals. This is false - only rhombi (and squares) have this property among parallelograms. Choice A is unnecessary since a rhombus is defined as having all sides congruent. Choice C incorrectly suggests that bisection is required for perpendicularity. Choice D suggests an unnecessarily complex approach when the theorem about rhombus diagonals should be applied directly.

Question 6

In triangle $ABC$ , $\overline{AD}$ is a median to side $\overline{BC}$ , and $\overline{BE}$ is a median to side $\overline{AC}$ . The medians intersect at point $G$ . To prove that $AG = \frac{2}{3}AD$ , which statement would be the most direct approach in a proof?

Show that $G$ is the centroid by proving that all three medians intersect at the same point, then apply the centroid theorem (correct answer)
Use the triangle inequality to establish that $AG$ must be less than $AD$ , then calculate the exact ratio using coordinate geometry
Prove that triangles $ABG$ and $CDG$ are congruent using SAS, then use corresponding parts to find the ratio
Apply the midpoint theorem to show that $\overline{AD}$ and $\overline{BE}$ bisect each other, then use properties of parallelograms

Explanation: The correct approach is to establish that G is the centroid (the intersection point of the three medians) and then apply the centroid theorem, which states that the centroid divides each median in a 2:1 ratio. Choice B incorrectly suggests using triangle inequality and coordinate geometry unnecessarily. Choice C mentions triangles that don't exist in this configuration (there is no point C mentioned as intersecting at G). Choice D incorrectly applies the midpoint theorem, which doesn't apply to medians intersecting.

Question 7

Quadrilateral $ABCD$ has $AB = CD$ and $AB \parallel CD$ . A student wants to prove that $ABCD$ is a parallelogram. Which statement identifies why this proof would be incomplete without additional information?

The definition of a parallelogram requires both pairs of opposite sides to be parallel, but only one pair is given as parallel in this problem
Having one pair of opposite sides both parallel and congruent is sufficient to prove a parallelogram, so no additional information is needed (correct answer)
The proof requires showing that all four sides are congruent, which cannot be determined from the given information about only two sides
The diagonals must be proven to bisect each other before concluding that the quadrilateral is a parallelogram, which requires coordinate geometry methods

Explanation: One of the standard theorems for proving a quadrilateral is a parallelogram states: 'If one pair of opposite sides is both parallel and congruent, then the quadrilateral is a parallelogram.' Since AB ∥ CD and AB = CD, this condition is satisfied, making the proof complete. Choice A incorrectly suggests more information is needed. Choice C incorrectly requires all sides to be congruent (that would be a rhombus). Choice D suggests an unnecessary approach using diagonals.

Question 8

Given that $\triangle PQR \sim \triangle STU$ with a ratio of similarity $2:3$ , and the area of $\triangle PQR$ is $16$ square units, a proof needs to establish the area of $\triangle STU$ . Which statement correctly describes the relationship needed?

Since the triangles are similar with ratio $2:3$ , the area ratio is also $2:3$ , so the area of $\triangle STU$ is $24$ square units
The area ratio equals the square of the similarity ratio, so the area of $\triangle STU$ is $16 \times \left(\frac{3}{2}\right)^2 = 36$ square units (correct answer)
Since corresponding sides are in the ratio $2:3$ , the area of $\triangle STU$ is $16 \times \frac{3}{2} = 24$ square units
The area relationship requires the cube of the similarity ratio for three-dimensional scaling, giving $16 \times \left(\frac{3}{2}\right)^3 = 54$ square units

Explanation: For similar figures, the ratio of areas equals the square of the ratio of corresponding linear dimensions. Since the similarity ratio is 2:3, the area ratio is (2:3)² = 4:9. Therefore, if triangle PQR has area 16, then triangle STU has area 16 × (3/2)² = 16 × 9/4 = 36. Choice A incorrectly uses the linear ratio for area. Choice C makes the same error. Choice D incorrectly applies a cubic relationship, which applies to volume, not area.

Question 9

Two circles with centers $O_1$ and $O_2$ are externally tangent at point $P$ . A common external tangent touches the circles at points $Q$ and $R$ respectively. To prove that $\overline{O_1Q} \parallel \overline{O_2R}$ , which property is essential to the proof?

Both $\overline{O_1Q}$ and $\overline{O_2R}$ are perpendicular to the common tangent line, making them parallel by the perpendicular transversal theorem (correct answer)
The circles have equal radii, so corresponding radii to tangent points must be parallel by the properties of congruent circles
Points $Q$ , $P$ , and $R$ are collinear, creating corresponding angles that make $\overline{O_1Q}$ and $\overline{O_2R}$ parallel
The common external tangent creates similar triangles $O_1PQ$ and $O_2PR$ , and corresponding sides of similar triangles are parallel

Explanation: The key property is that a radius to a point of tangency is always perpendicular to the tangent line. Since both O₁Q and O₂R are perpendicular to the same line (the common external tangent), they are parallel to each other. Choice B incorrectly assumes equal radii. Choice C incorrectly states that Q, P, and R are collinear. Choice D mentions triangles that don't exist in this configuration and misapplies similarity.

Question 10

Given that quadrilateral $PQRS$ has $\overline{PQ} \parallel \overline{SR}$ and $\overline{PS} \parallel \overline{QR}$ , a student wants to prove that $\overline{PQ} \cong \overline{SR}$ . Which sequence of statements represents a valid logical progression?

Since opposite sides are parallel, $PQRS$ is a parallelogram by definition; in a parallelogram, opposite sides are congruent by the parallelogram theorem (correct answer)
Draw diagonal $\overline{PR}$ ; since alternate interior angles are congruent when parallel lines are cut by a transversal, triangles $PQR$ and $SRP$ are congruent by ASA
Since $\overline{PQ} \parallel \overline{SR}$ , corresponding angles are congruent; therefore $\overline{PQ} \cong \overline{SR}$ by the converse of the isosceles triangle theorem
Apply the triangle inequality to triangles $PQS$ and $QRS$ ; since the sides are parallel, the triangles are similar and corresponding sides are proportional

Explanation: Choice A correctly identifies that having both pairs of opposite sides parallel makes PQRS a parallelogram by definition, and then applies the theorem that opposite sides of a parallelogram are congruent. Choice B correctly starts with drawing a diagonal but would need to prove the triangles are congruent first. Choice C incorrectly connects parallel sides to the isosceles triangle theorem. Choice D incorrectly applies triangle inequality and doesn't establish the necessary similarity.

Question 11

Given quadrilateral $PQRS$ where $\overline{PQ} \parallel \overline{SR}$ and $\overline{PS} \parallel$ QR $. To prove that$ PQRS$$ is a parallelogram using the definition of a parallelogram, what additional step is required after establishing the parallel sides?

Prove that opposite angles are congruent by showing $\angle P \cong \angle R$ and $\angle Q \cong \angle S$ using properties of parallel lines
Demonstrate that the diagonals bisect each other by proving $\overline{PR}$ and $\overline{QS}$ intersect at their midpoints
Show that opposite sides are congruent by proving $\overline{PQ} \cong \overline{SR}$ and $\overline{PS} \cong \overline{QR}$ using triangle congruence
No additional step is required since the definition of a parallelogram is satisfied by having both pairs of opposite sides parallel (correct answer)

Explanation: The correct answer is D. By definition, a parallelogram is a quadrilateral with both pairs of opposite sides parallel. Since we're given that

\overline{PQ} \parallel \overline{SR}

and

\overline{PS} \parallel \overline{QR}

, we have satisfied the definition completely. No additional steps are needed. Choice A describes a theorem about parallelograms but isn't part of the definition. Choice B describes another property of parallelograms but again isn't the definition. Choice C describes yet another property that can be proven once we know it's a parallelogram, but isn't required to prove it is one using the definition.

Question 12

Given that $\overline{MN} \parallel \overline{PQ}$ and transversal $\overline{RS}$ intersects both lines, with $\angle MTS = 3x + 20°$ and $\angle PUS = 5x - 40°$ where $T$ and $U$ are the intersection points. Which approach would verify that these angles are supplementary consecutive interior angles?

Set up the equation $3x + 20 + 5x - 40 = 180$ and solve for $x$ , then verify both angles are positive and their sum is $180°$
First prove that $\angle MTS$ and $\angle PUS$ are consecutive interior angles by their position, then apply the theorem that consecutive interior angles are supplementary (correct answer)
Show that $\angle MTS \cong \angle PUS$ by setting $3x + 20 = 5x - 40$ , then prove they are alternate interior angles which are congruent
Demonstrate that the angles are corresponding angles by their position relative to the parallel lines and transversal, then use the corresponding angles theorem

Explanation: The correct answer is B. To write a proper geometric proof, we must first establish that the angles are consecutive interior angles (also called co-interior or same-side interior angles) by examining their positions relative to the parallel lines and transversal. Once we've proven they are consecutive interior angles, we can apply the theorem that consecutive interior angles formed by parallel lines and a transversal are supplementary. Choice A jumps to calculation without proving the geometric relationship. Choice C incorrectly tries to prove the angles are congruent, which they're not if they're supplementary. Choice D misidentifies the angle relationship as corresponding angles.

Question 13

Two chords $\overline{AB}$ and $\overline{CD}$ intersect inside circle $O$ at point $P$ . Given that $AP = 4$ , $PB = 9$ , and $CP = 6$ , which statement correctly describes the logical flow for proving that $PD = 6$ ?

Apply the Intersecting Chords Theorem directly: since $AP \cdot PB = CP \cdot PD$ , substitute known values to get $4 \cdot 9 = 6 \cdot PD$ , then solve for $PD$
First prove that triangles $APC$ and $DPB$ are similar using inscribed angle relationships, then set up the proportion $\frac{AP}{DP} = \frac{CP}{BP}$ and solve
Show that $\angle APC \cong \angle DPB$ (vertical angles) and $\angle CAP \cong \angle BDP$ (inscribed angles subtending equal arcs), prove triangles $APC \sim DPB$ by AA, then use $AP \cdot PB = CP \cdot PD$ (correct answer)
Establish that the power of point $P$ with respect to circle $O$ can be calculated using both chords, leading to $AP \cdot PB = CP \cdot PD$ as equal expressions for this power

Explanation: The correct answer is C. A complete proof should establish why the Intersecting Chords Theorem works, not just apply it. We need to prove that triangles

APC

and

DPB

are similar by showing: (1)

\angle APC \cong \angle DPB

because they are vertical angles, and (2)

\angle CAP \cong \angle BDP

because they are inscribed angles that subtend the same arc. With two pairs of congruent angles (AA similarity), we can conclude the triangles are similar, which leads to the proportion that gives us the Intersecting Chords Theorem. Choice A applies the theorem without proving it. Choice B has the wrong proportion. Choice D mentions power of a point but doesn't develop the proof structure.

Question 14

Two circles with centers $O_1$ and $O_2$ intersect at points $P$ and $Q$ . If $\overline{O_1P} \cong \overline{O_1Q}$ and $\overline{O_2P} \cong \overline{O_2Q}$ , which statement best explains why $\overline{O_1O_2} \perp \overline{PQ}$ ?

Since both $O_1$ and $O_2$ are equidistant from $P$ and $Q$ , they both lie on the perpendicular bisector of $\overline{PQ}$ , making $\overline{O_1O_2}$ that perpendicular bisector (correct answer)
Triangle $O_1PQ$ is isosceles with $\overline{O_1P} \cong \overline{O_1Q}$ , and triangle $O_2PQ$ is isosceles with $\overline{O_2P} \cong \overline{O_2Q}$ , so corresponding altitudes are congruent
The intersection points of two circles always create perpendicular diameters, and since $\overline{O_1O_2}$ connects the centers, it must be perpendicular to $\overline{PQ}$
By the properties of circle intersection, the line connecting centers bisects the common chord at a right angle due to symmetry of circular arcs

Explanation: The correct answer is A. Since

O_1

is equidistant from

P

and

Q

(both are radii of the first circle),

O_1

lies on the perpendicular bisector of

\overline{PQ}

. Similarly, since

O_2

is equidistant from

P

and

Q

O_2

also lies on the perpendicular bisector of

\overline{PQ}

. Since two points determine a unique line,

\overline{O_1O_2}

must be the perpendicular bisector of

\overline{PQ}

. Choice B correctly identifies the isosceles triangles but incorrectly concludes about altitudes. Choice C makes a false claim about intersection points. Choice D uses vague reasoning without proper justification.

Question 15

In triangle $ABC$ , $\overline{DE} \parallel \overline{BC}$ where $D$ is on $\overline{AB}$ and $E$ is on $\overline{AC}$ . Given that $AD = 6$ , $DB = 4$ , and $AE = 9$ , what must be proven before applying the Triangle Proportionality Theorem to find $EC$ ?

Verify that triangles $ADE$ and $ABC$ are similar by confirming that $\angle A$ is shared and corresponding angles are equal due to parallel lines
Establish that $\overline{DE}$ divides sides $\overline{AB}$ and $\overline{AC}$ proportionally by confirming the parallel relationship creates similar triangles
Prove that point $D$ divides $\overline{AB}$ internally and point $E$ divides $\overline{AC}$ internally, ensuring the theorem applies to internal division
No additional proof is needed since $\overline{DE} \parallel \overline{BC}$ is given, which directly allows application of the Triangle Proportionality Theorem (correct answer)

Explanation: The correct answer is D. The Triangle Proportionality Theorem states that if a line is parallel to one side of a triangle and intersects the other two sides, then it divides those sides proportionally. Since we're given that

\overline{DE} \parallel \overline{BC}

and

D

and

E

are on sides

\overline{AB}

and

\overline{AC}

respectively, we can directly apply the theorem:

\frac{AD}{DB} = \frac{AE}{EC}

. Choice A describes proving similarity, which is related but not required to state before applying the theorem. Choice B essentially restates the theorem rather than identifying a prerequisite. Choice C mentions internal division, but this is evident from the given information and doesn't need separate proof.

Question 16

In rhombus $WXYZ$ , the diagonals $\overline{WY}$ and $\overline{XZ}$ intersect at point $M$ . To prove that $\triangle WXM \cong \triangle WZM$ , which property of rhombuses is essential but often overlooked in student proofs?

All sides of a rhombus are congruent, so $\overline{WX} \cong \overline{WZ}$ , combined with $\overline{WM} \cong \overline{WM}$ (reflexive) and the fact that diagonals bisect each other
Opposite angles in a rhombus are congruent, making $\angle XWY \cong \angle ZWY$ , which establishes the angle bisector property needed for the proof
The diagonals of a rhombus are perpendicular, creating right angles at $M$ , so $\angle WMX \cong \angle WMZ$ , which is necessary for angle-side-angle congruence (correct answer)
The diagonals bisect each other, so $\overline{XM} \cong \overline{ZM}$ , and combined with $\overline{WX} \cong \overline{WZ}$ and $\overline{WM} \cong \overline{WM}$ , we can use SSS congruence

Explanation: When proving triangle congruence in rhombuses, you need to identify which properties provide the specific angle or side relationships required for your chosen congruence theorem. The key insight here is recognizing that perpendicular diagonals create the angle congruence needed for a clean proof. In any rhombus, the diagonals intersect at right angles, making

\angle WMX \cong \angle WMZ

(both are 90°). Combined with

\overline{WX} \cong \overline{WZ}

(all sides of a rhombus are congruent) and

\overline{WM} \cong \overline{WM}

(reflexive property), you can use SAS congruence. This perpendicularity is what makes option C correct and often gets overlooked by students who focus on other rhombus properties. Option A mentions that diagonals bisect each other, but this property doesn't directly help prove

\triangle WXM \cong \triangle WZM

since the bisection creates equal segments on the opposite diagonal, not the sides of these specific triangles. Option B incorrectly states that opposite angles being congruent leads to angle bisection. While

\angle XWY \cong \angle ZWY

is true, this comes from the rhombus being a parallelogram, not from opposite angles being equal. Option D attempts SSS congruence but incorrectly uses

\overline{XM} \cong \overline{ZM}

. The diagonal bisection property gives us

\overline{WM} \cong \overline{YM}

and

\overline{XM} \cong \overline{ZM}

, but these don't form the sides of triangles

WXM

and

WZM

. Remember: when proving triangle congruence in rhombuses, the perpendicular diagonals often provide the most direct path to angle congruence theorems.

Question 17

In triangle $XYZ$ , the altitude from $Y$ to side $\overline{XZ}$ has length $h$ , and $\overline{XZ} = b$ . Triangle $X'Y'Z'$ is similar to triangle $XYZ$ with a scale factor of $k$ . To prove that the area of triangle $X'Y'Z'$ is $k^2$ times the area of triangle $XYZ$ , which relationship must be established?

The altitude from $Y'$ to side $\overline{X'Z'}$ has length $kh$ , and since $\overline{X'Z'} = kb$ , the area scales by $k \cdot k = k^2$ (correct answer)
All corresponding linear measurements scale by factor $k$ , so the area formula $A = \frac{1}{2} \cdot \text{base} \cdot \text{height}$ shows area scales by $k^2$
The ratio of areas equals the square of the ratio of corresponding sides, which is $\left(\frac{\text{side of } X'Y'Z'}{\text{side of } XYZ}\right)^2 = k^2$
Since the triangles are similar with scale factor $k$ , all area measurements are multiplied by $k^2$ by the definition of similarity ratio

Explanation: The correct answer is A. To prove the area relationship, we must show how both the base and height scale in similar triangles. Since triangle

X'Y'Z'

is similar to triangle

XYZ

with scale factor

k

, we have

\overline{X'Z'} = k \cdot \overline{XZ} = kb

. The altitude from

Y'

\overline{X'Z'}

kh

because all linear measurements scale by

k

. Therefore, Area of

X'Y'Z' = \frac{1}{2}(kb)(kh) = k^2 \cdot \frac{1}{2}bh = k^2 \cdot \text{Area of } XYZ

. Choice B states the conclusion without showing the scaling relationship. Choice C uses a theorem without proving it. Choice D incorrectly treats the

k^2

relationship as a definition rather than something to be proven.

Question 18

In triangle $ABC$ , point $D$ is on side $\overline{BC}$ such that $\angle BAD = \angle CAD$ . Given that $AB = 8$ , $AC = 6$ , and $BC = 10$ , which statement provides the most direct approach to prove that $\frac{BD}{DC} = \frac{AB}{AC}$ ?

Use the Law of Cosines on triangles $ABD$ and $ACD$ to establish that $AD$ bisects $\angle BAC$ , then apply the Angle Bisector Theorem
Apply the Angle Bisector Theorem directly since $\overline{AD}$ bisects $\angle BAC$ by the given condition $\angle BAD = \angle CAD$ (correct answer)
Prove that triangles $ABD$ and $ACD$ are similar using AA similarity, then use the ratio of corresponding sides
Show that $\overline{AD}$ is perpendicular to $\overline{BC}$ using the given angle condition, then apply properties of right triangles

Explanation: The correct answer is B. Since we're given that

\angle BAD = \angle CAD

, this directly tells us that

\overline{AD}

bisects

\angle BAC

. The Angle Bisector Theorem states that if a ray bisects an angle of a triangle, then it divides the opposite side into segments whose lengths are proportional to the adjacent sides. Therefore,

\frac{BD}{DC} = \frac{AB}{AC}

. Choice A is unnecessarily complex since we already know

AD

bisects the angle. Choice C is incorrect because triangles

ABD

and

ACD

share angle

A

but are not similar. Choice D is wrong because angle bisection doesn't imply perpendicularity.

Question 19

Triangle $DEF$ has vertices $D(0, 0)$ , $E(6, 0)$ , and $F(3, 4)$ . To prove that the triangle is isosceles, which calculation sequence provides the most direct approach?

Find the slopes of all three sides and verify that two sides have equal slopes, indicating that two sides are parallel and therefore equal
Calculate $DE = \sqrt{(6-0)^2 + (0-0)^2} = 6$ , $DF = \sqrt{(3-0)^2 + (4-0)^2} = 5$ , and $EF = \sqrt{(3-6)^2 + (4-0)^2} = 5$ (correct answer)
Use the midpoint formula to find the midpoints of all sides, then verify that one median equals half the length of its corresponding side
Apply the angle bisector theorem by calculating the angles at each vertex using the dot product formula and verifying that two angles are equal

Explanation: To prove a triangle is isosceles, we need to show that two sides have equal length. Choice B correctly calculates all three side lengths using the distance formula and shows that DF = EF = 5, proving the triangle is isosceles. Choice A incorrectly suggests that parallel sides would make sides equal (which is impossible in a triangle). Choice C applies an irrelevant property about medians. Choice D is unnecessarily complex when direct distance calculation is simpler.