Bisector Proofs - Math 2

1. Ray $QS$ bisects $\angle PQR$. If $\triangle PQS \cong \triangle RQS$, which additional piece of information would allow you to conclude that $QS$ is also the perpendicular bisector of $\overline{PR}$?
   Answer: $\angle PSQ = \angle RSQ = 90°$ and $QS$ intersects $\overline{PR}$ Since $\triangle PQS \cong \triangle RQS$, we have $PS = RS$, so $S$ is equidistant from $P$ and $R$. For $QS$ to be the perpendicular bisector of $\overline{PR}$, we need: (1) $S$ lies on $\overline{PR}$, and (2) $QS \perp PR$. Option A provides both conditions directly. Option B doesn't ensure $QS \perp PR$. Option C restates given information. Option D gives that $\angle PSQ$ and $\angle RSQ$ are supplementary, but without $S$ on $\overline{PR}$, this doesn't guarantee they're both $90°$.
2. Triangle $STU$ has an incenter at point $I$. The angle bisector from vertex $S$ intersects side $TU$ at point $V$, and the perpendicular from $I$ to side $TU$ intersects $TU$ at point $W$. Which statement about the relationship between $SV$ and $IW$ is most accurate?
   Answer: $SV$ contains point $I$, and $IW$ is perpendicular to $TU$, but $SV$ is not necessarily the perpendicular bisector of $TU$ Since $I$ is the incenter, it lies on all angle bisectors of the triangle, including the angle bisector $SV$ from vertex $S$. The perpendicular $IW$ from the incenter to side $TU$ is indeed perpendicular to $TU$ (this is how the inradius is defined). However, $SV$ is only the perpendicular bisector of $TU$ when triangle $STU$ is isosceles with $ST = SU$. In a general triangle, the angle bisector from $S$ does not bisect the opposite side unless the triangle is isosceles. Choice A incorrectly suggests they're the same segment. Choice C correctly identifies the isosceles condition but incorrectly states it as the main relationship. Choice D incorrectly claims they always form a perpendicular bisector.
3. In rhombus $WXYZ$, the diagonals $WY$ and $XZ$ intersect at point $O$. A student claims that $WY$ is the perpendicular bisector of $XZ$. Which reasoning best justifies this claim?
   Answer: Since $WX = WZ$ and $YX = YZ$, both $W$ and $Y$ are equidistant from $X$ and $Z$, so $WY$ is the perpendicular bisector In a rhombus, all sides are equal, so $WX = WZ$ and $YX = YZ$. This means both $W$ and $Y$ are equidistant from $X$ and $Z$, placing both points on the perpendicular bisector of $XZ$. Since two points determine a unique line, $WY$ must be the perpendicular bisector of $XZ$. Choice A is too general without showing the reasoning. Choice B incorrectly identifies the congruence as SSS when it should be SAS. Choice D correctly states properties but doesn't use the equidistant reasoning that directly establishes the perpendicular bisector.
4. Given quadrilateral $PQRS$ with diagonal $PR$, suppose $PQ = PS$ and $QR = SR$. To prove that $PR$ is the perpendicular bisector of $QS$, which congruence statement provides the most direct justification?
   Answer: $\triangle PQR \cong \triangle PSR$ by SSS, which implies $\angle QPR = \angle SPR$ To prove $PR$ is the perpendicular bisector of $QS$, we need to show that $PR$ bisects $QS$ at right angles. Using $\triangle PQR \cong \triangle PSR$ by SSS (given $PQ = PS$, $QR = SR$, $PR = PR$), we get $\angle QPR = \angle SPR$, so $PR$ bisects $\angle QPS$. Combined with $PQ = PS$, this makes $P$ equidistant from $Q$ and $S$, placing $P$ on the perpendicular bisector of $QS$. Since any two points determine a unique line, $PR$ must be that perpendicular bisector. Choice B uses an invalid triangle. Choice C assumes the intersection point exists. Choice D doesn't lead to the angle bisector property needed.
5. In isosceles triangle $DEF$ with $DE = DF$, the altitude from $D$ to side $EF$ intersects $EF$ at point $G$. To prove that $DG$ is the perpendicular bisector of $EF$, which congruence statement is most direct?
   Answer: $\triangle DEG \cong \triangle DFG$ by HL using $DE = DF$ as hypotenuse Since $DG$ is the altitude to $EF$, we have $\angle DGE = \angle DGF = 90°$, making both triangles right triangles. We can use HL (Hypotenuse-Leg): $DE = DF$ (hypotenuses, given) and $DG = DG$ (shared leg). This proves $\triangle DEG \cong \triangle DFG$, giving us $EG = FG$, so $G$ bisects $EF$. Combined with $DG \perp EF$, $DG$ is the perpendicular bisector. While other methods work, HL is most direct for right triangles.
6. Consider triangle $XYZ$ where $\angle X = 40°$, $\angle Y = 70°$, and $\angle Z = 70°$. If point $W$ is on side $YZ$ such that $XW$ bisects $\angle YXZ$, what additional property does $XW$ possess?
   Answer: $XW$ is simultaneously the altitude, median, and perpendicular bisector of $YZ$ Since $\angle Y = \angle Z = 70°$, triangle $XYZ$ is isosceles with $XY = XZ$. In an isosceles triangle, the angle bisector from the vertex angle ($\angle X$) is also the altitude, median, and perpendicular bisector of the base. Since $XW$ bisects $\angle YXZ$, it is perpendicular to $YZ$, bisects $YZ$ (making $W$ the midpoint), and is therefore the perpendicular bisector of $YZ$.
7. Point $P$ lies on the perpendicular bisector of segment $XY$. If $PX = 3k + 7$ and $PY = 5k - 1$ where $k > 0$, what is the value of $k$?
   Answer: $k = 4$ Since $P$ lies on the perpendicular bisector of $XY$, by the definition of perpendicular bisector, $P$ is equidistant from $X$ and $Y$. Therefore $PX = PY$. Setting up the equation: $3k + 7 = 5k - 1$. Solving: $7 + 1 = 5k - 3k$, so $8 = 2k$, giving $k = 4$. We can verify: $PX = 3(4) + 7 = 19$ and $PY = 5(4) - 1 = 19$.
8. Given that $M$ is the midpoint of segment $PQ$, and line $\ell$ passes through $M$ such that $\angle PM\ell = 90°$. Which additional condition is sufficient to prove that $\ell$ is the perpendicular bisector of $PQ$?
   Answer: Any point on $\ell$ is equidistant from $P$ and $Q$ Since $M$ is the midpoint of $PQ$ and $\ell \perp PQ$ at $M$, we have the perpendicular through the midpoint. To complete the proof that $\ell$ is the perpendicular bisector, we need to show the defining property: every point on $\ell$ is equidistant from $P$ and $Q$. Option B is automatically true for any perpendicular. Option C is irrelevant to the bisector property. Option D is also automatically satisfied.
9. Points $J$, $K$, and $L$ are positioned such that $JK = JL$ and $MK = ML$, where $M$ is a point not on line $JK$ or $JL$. If $\angle KJM = \angle LJM$, what is the most complete conclusion about line $JM$?
   Answer: $JM$ is the angle bisector of $\angle KJL$ and perpendicular bisector of $KL$ Given $\angle KJM = \angle LJM$, $JM$ bisects $\angle KJL$. We can prove $\triangle JMK \cong \triangle JML$ by SAS: $JK = JL$ (given), $\angle KJM = \angle LJM$ (given), and $JM = JM$ (reflexive). From congruence, $MK = ML$, which confirms the given condition. Since both $J$ and $M$ are equidistant from $K$ and $L$, line $JM$ is the perpendicular bisector of $KL$ by the converse of the perpendicular bisector theorem.
10. In rhombus $ABCD$, diagonal $AC$ intersects diagonal $BD$ at point $E$. To prove that $AC$ is the perpendicular bisector of $BD$, which property of rhombuses provides the most direct justification?
    Answer: Adjacent sides are congruent, making triangles $\triangle ABE \cong \triangle CBE$ by SSS congruence In rhombus $ABCD$, all sides are congruent: $AB = BC = CD = DA$. To prove $AC$ is the perpendicular bisector of $BD$, we prove $\triangle ABE \cong \triangle CBE$ by SSS: $AB = CB$ (sides of rhombus), $AE = CE$ (diagonals of rhombus bisect each other), and $BE = BE$ (reflexive). From congruence, $\angle AEB = \angle CEB$. Since these angles are supplementary and congruent, each is $90°$, proving $AC \perp BD$. Combined with $E$ being the midpoint, $AC$ is the perpendicular bisector of $BD$.
11. In the coordinate plane, line $\ell$ passes through points $A(2, 5)$ and $B(8, 5)$. Point $C$ is located at $(5, 2)$. Which statement about the relationship between point $C$ and segment $AB$ can be verified?
    Answer: $C$ lies on the perpendicular bisector of $AB$ because it's equidistant from $A$ and $B$ First, find distances: $CA = \sqrt{(5-2)^2 + (2-5)^2} = \sqrt{9 + 9} = 3\sqrt{2}$ and $CB = \sqrt{(5-8)^2 + (2-5)^2} = \sqrt{9 + 9} = 3\sqrt{2}$. Since $CA = CB$, point $C$ is equidistant from $A$ and $B$, so it lies on the perpendicular bisector of $AB$. The midpoint of $AB$ is $(5, 5)$, and the perpendicular bisector is the vertical line $x = 5$. Indeed, $C(5, 2)$ lies on this line.
12. Two circles intersect at points $M$ and $N$. Point $P$ is the center of one circle and point $Q$ is the center of the other circle. Based on the properties of circles, what can be concluded about line $PQ$ in relation to chord $MN$?
    Answer: Line $PQ$ is the perpendicular bisector of $MN$ because both centers are equidistant from $M$ and $N$ Since $M$ and $N$ lie on the circle centered at $P$, we have $PM = PN$ (both equal to the radius of circle $P$). Similarly, since $M$ and $N$ lie on the circle centered at $Q$, we have $QM = QN$ (both equal to the radius of circle $Q$). Since both $P$ and $Q$ are equidistant from $M$ and $N$, both points lie on the perpendicular bisector of $MN$. Since two points determine a unique line, line $PQ$ is the perpendicular bisector of $MN$.
13. In triangle $ABC$, point $D$ lies on side $BC$ such that $\angle BAD = \angle CAD$. If it is also given that $AB = AC$, which statement must be true?
    Answer: $AD$ is the angle bisector of $\angle BAC$ and the perpendicular bisector of $BC$ Since $AB = AC$ (given) and $\angle BAD = \angle CAD$ (given), triangles $ABD$ and $ACD$ are congruent by SAS ($AB = AC$, $\angle BAD = \angle CAD$, $AD = AD$). Therefore $BD = CD$ and $\angle ADB = \angle ADC$. Since these angles are supplementary and equal, each must be $90°$. Thus $AD$ both bisects $BC$ and is perpendicular to it, making it the perpendicular bisector. Choice A omits that $AD$ is also an angle bisector. Choice C incorrectly suggests $AD$ might not be the perpendicular bisector. Choice D incorrectly suggests $AD$ might not be perpendicular to $BC$.
14. Line segment $PQ$ has endpoints $P(-3, 1)$ and $Q(5, 7)$. Point $R$ lies on the perpendicular bisector of $PQ$. If $R$ has coordinates $(a, 4)$, what is the value of $a$?
    Answer: $a = 1$ Since $R$ lies on the perpendicular bisector of $PQ$, we have $RP = RQ$. Using the distance formula: $RP = \sqrt{(a-(-3))^2 + (4-1)^2} = \sqrt{(a+3)^2 + 9}$ and $RQ = \sqrt{(a-5)^2 + (4-7)^2} = \sqrt{(a-5)^2 + 9}$. Setting $RP = RQ$: $\sqrt{(a+3)^2 + 9} = \sqrt{(a-5)^2 + 9}$. Squaring both sides: $(a+3)^2 = (a-5)^2$. Expanding: $a^2 + 6a + 9 = a^2 - 10a + 25$. Simplifying: $16a = 16$, so $a = 1$.
15. In triangle $RST$, if $RU = SU$, $TU = TU$, and $\angle RUT = \angle SUT = 90°$, what can be concluded about line $TU$?
    Answer: $TU$ is both the perpendicular bisector of $RS$ and the angle bisector of $\angle RTS$ Given $RU = SU$ and both $\angle RUT = \angle SUT = 90°$, we can prove $\triangle RUT \cong \triangle SUT$ by SAS ($RU = SU$, $\angle RUT = \angle SUT$, $TU = TU$). From congruence: $TR = TS$ and $\angle RTU = \angle STU$. Since $\angle RTU = \angle STU$, $TU$ bisects $\angle RTS$. Since $RU = SU$ and $TU \perp RS$, $TU$ is the perpendicular bisector of $RS$.
16. Points $A$, $B$, and $C$ are positioned such that $A$ is equidistant from $B$ and $C$. Point $D$ is also equidistant from $B$ and $C$, but $D \neq A$. If line $\ell$ passes through both $A$ and $D$, which additional condition is needed to prove that $\ell$ is the perpendicular bisector of $\overline{BC}$?
    Answer: No additional condition is needed; $\ell$ is necessarily the perpendicular bisector of $\overline{BC}$ Since both $A$ and $D$ are equidistant from $B$ and $C$, both points lie on the perpendicular bisector of $\overline{BC}$. The perpendicular bisector of a line segment is the unique line consisting of all points equidistant from the endpoints of the segment. Since $A$ and $D$ are two distinct points on this line, and two points determine a unique line, line $\ell$ through $A$ and $D$ must be the perpendicular bisector of $\overline{BC}$. Choice A is automatically satisfied. Choice B is not necessary. Choice C describes a coincidental relationship that's not required for the perpendicular bisector property.
17. Line $m$ passes through point $M$, the midpoint of $\overline{AB}$, and makes a $90°$ angle with $\overline{AB}$. Point $C$ is chosen on line $m$ such that $C \neq M$. Which method would best prove that $m$ is the perpendicular bisector of $\overline{AB}$?
    Answer: Show that $\triangle AMC \cong \triangle BMC$ using SAS congruence postulate To prove $m$ is the perpendicular bisector, we need to show any point on $m$ is equidistant from $A$ and $B$. Using SAS: $AM = BM$ ($M$ is midpoint), $\angle AMC = \angle BMC = 90°$ ($m \perp AB$), and $MC = MC$ (reflexive). This proves $\triangle AMC \cong \triangle BMC$, so $AC = BC$. Option A can't work since we don't know $AC = BC$ initially. Option C only shows the angles are equal, not the full bisector property. Option D is irrelevant.
18. In triangle $DEF$, point $G$ lies on side $EF$ such that $DG \perp EF$. Additionally, $\angle EDG = \angle FDG$. What can be concluded about the relationship between $DG$ and side $EF$?
    Answer: $DG$ is both the altitude and angle bisector, which implies that triangle $DEF$ is isosceles with $DE = DF$ Given that $DG \perp EF$ (altitude condition) and $\angle EDG = \angle FDG$ (angle bisector condition), we have a line from vertex $D$ that serves both as altitude and angle bisector to side $EF$. In any triangle, when the altitude and angle bisector from the same vertex to the opposite side are the same line, the triangle must be isosceles with the two sides adjacent to that vertex being equal. Therefore $DE = DF$. This also makes $DG$ the perpendicular bisector of $EF$ since $G$ is equidistant from $E$ and $F$. Choice A ignores the angle bisector property. Choice B ignores the altitude property. Choice D correctly identifies the perpendicular bisector but doesn't capture the key insight about the isosceles triangle.
19. In the coordinate plane, points $P(2, 5)$, $Q(-1, 3)$, and $R(3, 1)$ form a triangle. If line $s$ passes through $P$ and is the perpendicular bisector of $\overline{QR}$, what must be true about the coordinates of any point on line $s$?
    Answer: Every point $(x, y)$ on line $s$ satisfies $(x + 1)^2 + (y - 3)^2 = (x - 3)^2 + (y - 1)^2$ For line $s$ to be the perpendicular bisector of $\overline{QR}$, every point on $s$ must be equidistant from $Q(-1, 3)$ and $R(3, 1)$. This means that for any point $(x, y)$ on line $s$, the distance to $Q$ equals the distance to $R$: $\sqrt{(x + 1)^2 + (y - 3)^2} = \sqrt{(x - 3)^2 + (y - 1)^2}$. Squaring both sides gives $(x + 1)^2 + (y - 3)^2 = (x - 3)^2 + (y - 1)^2$. We can verify that $P(2, 5)$ satisfies this equation. Choice B uses the wrong points. Choice C gives a specific linear equation but doesn't capture the equidistant property. Choice D has an incorrect linear equation.

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