Math 2 Quiz: Critiquing Proofs

A proof about circle tangent properties states: "Line $l$ is tangent to circle $O$ at point $P$. Since tangent lines are perpendicular to radii at the point of tangency, $l \perp OP$. Therefore, $\angle OPQ = 90°$ for any point $Q$ on line $l$." What error appears in this reasoning?

A student attempts to prove that quadrilateral $JKLM$ is a rhombus: "Since opposite sides are parallel ($JK \parallel LM$ and $JM \parallel KL$), $JKLM$ is a parallelogram. Since all sides are congruent ($JK = KL = LM = MJ$), $JKLM$ is a rhombus." Which aspect of this proof requires the most scrutiny?

Examine this proof that the sum of interior angles of quadrilateral $WXYZ$ is $360°$: "Drawing diagonal $WY$ divides the quadrilateral into triangles $WXY$ and $WYZ$. Since each triangle has angle sum $180°$, the total is $180° + 180° = 360°$." What critical verification is missing?

A proof states: "In right triangle $XYZ$ with right angle at $Y$, if $XY = 3$ and $YZ = 4$, then by the Pythagorean Theorem, $XZ^2 = XY^2 + YZ^2 = 3^2 + 4^2 = 9 + 16 = 25$. Therefore $XZ = 5$ and the triangle satisfies the Pythagorean Theorem." What logical issue exists in this reasoning?

Consider this proof involving similar triangles: "Triangles $ABC$ and $DEF$ have $\angle A = \angle D$, $\angle B = \angle E$, and $\angle C = \angle F$. By AAA similarity, $\triangle ABC \sim \triangle DEF$. Therefore, corresponding sides are proportional: $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$." What issue affects the validity of this proof?

In the coordinate proof shown below, what assumption needs explicit justification? "To prove that quadrilateral $ABCD$ with vertices $A(0,0)$, $B(4,0)$, $C(4,3)$, $D(0,3)$ is a rectangle, we calculate: $AB = 4$, $BC = 3$, $CD = 4$, $DA = 3$. Since opposite sides are equal, $ABCD$ is a parallelogram. Since $AB \perp BC$ (slopes are $0$ and undefined), all angles are $90°$, making $ABCD$ a rectangle."

A student proves that the diagonals of rectangle $ABCD$ are congruent using this logic: "Since $ABCD$ is a rectangle, all angles are $90°$. In triangles $ABC$ and $ABD$, we have $AB = AB$ (reflexive), $\angle ABC = \angle ABD = 90°$, and $BC = AD$ (opposite sides of rectangle). By SAS, $\triangle ABC \cong \triangle ABD$, so $AC = BD$." What is wrong with this proof?

Analyze this proof about the exterior angle of a triangle: "In triangle $DEF$, exterior angle $\angle EFG$ is formed by extending side $EF$ through $F$. Since $\angle EFG$ and $\angle EFD$ are supplementary, and $\angle D + \angle E + \angle EFD = 180°$ (Triangle Angle Sum), we get $\angle EFG = \angle D + \angle E$." What step in the reasoning needs clarification?

In the proof shown below, which step contains an unjustified leap in logic? "Given that $ABCD$ is a parallelogram with diagonals $AC$ and $BD$ intersecting at point $E$. Since opposite sides of a parallelogram are parallel and congruent, $AB \parallel DC$ and $AB = DC$. In triangles $ABE$ and $CDE$, we have $AB = DC$, $\angle BAE = \angle DCE$ (alternate interior angles), and $\angle ABE = \angle CDE$ (alternate interior angles). By ASA, $\triangle ABE \cong \triangle CDE$, so $AE = CE$ and $BE = DE$."

In the proof shown below, what critical step is missing between statements 3 and 4?

1. Given: $\overline{AB} \perp \overline{CD}$ at point $E$
2. $\angle AEC = 90°$ (Definition of perpendicular lines)
3. $\angle AED = 90°$ (Linear pair with $\angle AEC$)
4. $\overline{AB} \perp \overline{ED}$ (Definition of perpendicular lines)

A proof attempts to show that the diagonals of a rhombus are perpendicular. The proof states:

1. Let PQRS be a rhombus with diagonals PR and QS intersecting at point T

2. Since PQRS is a rhombus, all sides are equal: $PQ = QR = RS = SP$

3. Since all sides are equal, triangles PQT and RST are congruent by SSS

4. Therefore, $\angle PTQ = \angle RTS$

5. Since these angles are vertical angles, $\angle PTQ = \angle RTS = 90°$

6. Thus, the diagonals are perpendicular

Which step contains the most significant error?

Consider this proof that the perpendicular bisector of a chord passes through the center of a circle:

1. Let chord AB be in circle O, with M as the midpoint of AB

2. Let line ℓ be perpendicular to AB at point M

3. Since M is the midpoint, $AM = MB$

4. For any point P on line ℓ, triangles PAM and PBM are congruent by SAS

5. Therefore $PA = PB$ for any point P on line ℓ

6. Since the center O is equidistant from all points on the circle, $OA = OB$

7. By step 5, since O is equidistant from A and B, point O must lie on line ℓ

8. Therefore, the perpendicular bisector passes through the center

Which step contains flawed reasoning?

A student attempts to prove that the base angles of an isosceles triangle are equal using the following approach:

1. Given: Triangle ABC with $AB = AC$

2. Construct the angle bisector of $\angle BAC$, meeting BC at point D

3. Since AD bisects $\angle BAC$, we have $\angle BAD = \angle CAD$

4. In triangles ABD and ACD: $AB = AC$ (given), $\angle BAD = \angle CAD$ (from step 3), and $AD = AD$ (reflexive property)

5. Therefore, triangles ABD and ACD are congruent by SAS

6. From this congruence, $\angle ABD = \angle ACD$

7. Since $\angle ABD = \angle ABC$ and $\angle ACD = \angle ACB$, we have $\angle ABC = \angle ACB$

What issue affects the validity of this proof?

Examine this proof that opposite angles in a cyclic quadrilateral are supplementary:

1. Let PQRS be a cyclic quadrilateral inscribed in circle O

2. Draw radii OP, OQ, OR, and OS

3. The central angle $\angle POR$ corresponds to arc PR

4. The inscribed angle $\angle PQR$ also corresponds to arc PR

5. By the inscribed angle theorem, $\angle PQR = \frac{1}{2} \angle POR$

6. Similarly, $\angle PSR = \frac{1}{2} \angle POR$

7. Therefore, $\angle PQR = \angle PSR$

8. Since the sum of angles in quadrilateral PQRS is 360°, and $\angle PQR = \angle PSR$, we have $\angle PQR + \angle PSR = 180°$

Which step contains the most significant error?

A proof attempts to show that the medians of a triangle are concurrent (meet at a single point). The proof uses coordinate geometry:

1. Place triangle ABC with A(0, 0), B(2a, 0), and C(2b, 2c) for arbitrary values a, b, c

2. Find midpoints: M of BC is (a+b, c), N of AC is (b, c), P of AB is (a, 0)

3. Median from A to M has equation: $y = \frac{c}{a+b}x$

4. Median from B to N has equation: $y = \frac{c}{b-a}(x-2a)$

5. Setting these equal to find intersection: $\frac{c}{a+b}x = \frac{c}{b-a}(x-2a)$

6. Solving: $\frac{x}{a+b} = \frac{x-2a}{b-a}$

7. Cross-multiplying: $x(b-a) = (x-2a)(a+b)$

8. This gives $x = \frac{2a(a+b)}{3a+b}$

What assumption limits the generality of this proof?

A proof claims to show that if a quadrilateral has one pair of parallel sides, then it is a parallelogram. The proof states:

1. Given: Quadrilateral WXYZ with $WX \parallel YZ$

2. Draw diagonal WY

3. $\angle XWY = \angle ZYW$ (alternate interior angles with WX || YZ)

4. $\angle XYW = \angle ZWY$ (alternate interior angles with WX || YZ)

5. In triangles WXY and YZW: $\angle XWY = \angle ZYW$ and $\angle XYW = \angle ZWY$

6. Therefore, triangles WXY and YZW are congruent by AA

7. From congruence: $WX = YZ$ and $XY = ZW$

8. Since both pairs of opposite sides are equal, WXYZ is a parallelogram

What is wrong with this proof?

A student proves that if two chords in a circle are equidistant from the center, then they are equal in length:

1. Let AB and CD be chords in circle O, equidistant from center O

2. Let OM ⊥ AB at M, and ON ⊥ CD at N, with OM = ON

3. Since OM ⊥ AB, point M bisects chord AB, so AM = MB

4. Since ON ⊥ CD, point N bisects chord CD, so CN = ND

5. In right triangle OMA: $OA^2 = OM^2 + AM^2$

6. In right triangle ONC: $OC^2 = ON^2 + CN^2$

7. Since OA = OC (radii) and OM = ON (given), we have $AM^2 = CN^2$

8. Therefore AM = CN, and since AB = 2·AM and CD = 2·CN, we get AB = CD

Which step makes an unjustified assumption?

Consider this proof that an exterior angle of a triangle equals the sum of the two non-adjacent interior angles:

1. In triangle ABC, extend side BC to point D, creating exterior angle ACD

2. The exterior angle ACD and interior angle ACB are supplementary: $\angle ACD + \angle ACB = 180°$

3. The sum of interior angles in triangle ABC: $\angle BAC + \angle ABC + \angle ACB = 180°$

4. From steps 2 and 3: $\angle ACD + \angle ACB = \angle BAC + \angle ABC + \angle ACB$

5. Subtracting $\angle ACB$ from both sides: $\angle ACD = \angle BAC + \angle ABC$

6. Therefore, the exterior angle equals the sum of the two non-adjacent interior angles

Which aspect of this proof should be questioned?