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5.6 — Justification and Proof
In Section 5.5 you identified angle patterns and made conjectures. Today you will explain why those claims must hold. A convincing proof does not depend on how accurate the picture looks; it proceeds from given facts, definitions, and previously established properties.
1. In class: A Practice Understanding Task
From a conjecture to a proof
A proof is a chain of statements in which every important claim has a reason. Start by identifying what is given, what you must prove, and which previously known relationships connect them. Section 5.3 introduced narrative, flow, and two-column formats; all three can express the same reasoning.
Mark the given information; name the target claim; use definitions and previously established facts; state the conclusion only after the necessary intermediate steps. Reasons might be “given,” “linear-pair angles sum to 180°,” “triangle angle sum,” “translation preserves angles,” or “vertical angles are congruent” after that fact is proven.
Given: ∠A and ∠B are supplementary and have equal measures. Prove: both measure 90°.
| Statement | Reason |
|---|---|
| \(m\angle A+m\angle B=180^\circ\) | Definition of supplementary angles |
| \(m\angle A=m\angle B\) | Given |
| \(2m\angle A=180^\circ\) | Substitution |
| \(m\angle A=m\angle B=90^\circ\) | Divide by 2 and use equality |
Prove that vertical angles are congruent (main task 1)
When two lines intersect, opposite angles are vertical angles. An angle next to either vertical angle makes a linear pair with it. The key is that the two vertical angles are both supplements of the same adjacent angle.
Given: \(\overleftrightarrow{AC}\) and \(\overleftrightarrow{BD}\) intersect at E (the diagram labels have been arranged so both pairs of opposite rays lie on straight lines). Prove: \(\angle AEB\cong\angle CED\).
Both ∠AEB and ∠CED are supplementary to ∠BEC because they each form a linear pair with ∠BEC.
\[m\angle AEB+m\angle BEC=180^\circ,\]\[m\angle CED+m\angle BEC=180^\circ.\]Subtract the shared \(m\angle BEC\) from both equations:
\[\boxed{m\angle AEB=m\angle CED}.\]Angles with equal measures are congruent.
Our reconstructed diagram uses points A, E, C on one line and B, E, D on the other. The opposite-angle relationship is the essential given fact.
Prove the Exterior Angle Theorem (main task 2)
Extend one side of a triangle. The exterior angle and the adjacent interior angle form a linear pair. Meanwhile, the three interior angles of the triangle total 180°.
Given: angle 4 is exterior to a triangle. Prove: \(m\angle4=m\angle1+m\angle2\).
| Statement | Reason |
|---|---|
| \(m\angle1+m\angle2+m\angle3=180^\circ\) | Triangle interior-angle sum |
| \(m\angle3+m\angle4=180^\circ\) | Angles 3 and 4 form a linear pair |
| \(m\angle1+m\angle2+m\angle3=m\angle3+m\angle4\) | Both expressions equal 180° |
| \(\boxed{m\angle4=m\angle1+m\angle2}\) | Subtract \(m\angle3\) |
The two remote interior angles add to the exterior angle. The adjacent interior angle is not one of the remote angles.
Prove angle relationships for parallel lines (main tasks 3–5)
Let two horizontal lines be parallel and let a diagonal transversal cross them. The numbered positions below match the packet's principal relationships: corresponding 1 and 5, alternate interior 3 and 6, and same-side interior 3 and 5.
Corresponding angles (packet task 3)
Claim: angles 1 and 5 are congruent when the lines are parallel.
Move the first intersection to the second by a translation along the transversal. Because the transversal passes through both intersection points, that translation carries the transversal onto itself. It also carries the top line onto the unique parallel line through the second intersection: the bottom line. A translation preserves angle measure, so the angle in position 1 lands on the angle in matching position 5.
\[\boxed{\angle1\cong\angle5.}\]This is a rigid-motion justification, not a claim based on appearance.
Alternate interior angles (packet task 4)
To prove the alternate interior pair 3 and 6, connect two earlier results:
Angles 3 and 2 are vertical, so \(m\angle3=m\angle2\). Angles 2 and 6 are corresponding, so \(m\angle2=m\angle6\).
\[\boxed{m\angle3=m\angle6}.\]Same-side interior angles (packet task 5)
Angles 3 and 6 are alternate interior and therefore congruent. Angles 5 and 6 form a linear pair at the lower crossing, so their measures sum to 180°. Substituting the equal measures gives
\[m\angle5+m\angle6=180^\circ\]\[\boxed{m\angle5+m\angle3=180^\circ}.\]Therefore angles 3 and 5 are supplementary.
Reverse the implication: the converse (main task 6)
So far we assumed two lines were parallel and deduced equal angle measures. The converse reverses the condition and conclusion: if a transversal makes equal alternate interior angles, the two lines must be parallel.
Given: \(\angle3\cong\angle6\). Prove: \(\ell\parallel m\).
Suppose instead that a line \(m'\) through the lower intersection is drawn parallel to \(\ell\). The proved alternate-interior-angle property would make the matching angle on \(m'\) congruent to angle 3. By the given, the matching angle on the actual line \(m\) is also congruent to angle 3. Through a fixed point, there is only one ray on the indicated side of a transversal making a specified angle with it. Thus \(m=m'\); consequently \(\boxed{\ell\parallel m}\).
Key distinction: this argument relies on the uniqueness of a ray forming a given angle and the existence of a parallel through a point; simply reversing the earlier proof is not a valid logical step without a reason.
2. Ready — Recall rigid motions (problems 1–4)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
A rigid motion preserves distances and angles. All translations, reflections, and rotations are rigid motions, though their descriptions of corresponding points differ.
| Motion | What happens to corresponding points? | Why? |
|---|---|---|
| Translation | The segment joining each point to its image has the same length and is parallel to every other such joining segment (unless the displacement is zero). | Every point moves by the same vector. |
| Reflection | The reflecting line is the perpendicular bisector of a segment joining a point to its distinct image. | Mirror points are equally distant from the mirror line on opposite sides. |
| Rotation | A point and its image are the same distance from the rotation center. | The rotation moves each point along a circle centered at the rotation point. |
| Any rigid motion | Lengths and angle measures remain equal. | Rigid motions are distance-preserving isometries. |
Translate by the vector \(\langle 3,-2\rangle\). The point \((1,4)\) moves to \((4,2)\), while \((-2,1)\) moves to \((1,-1)\). Each joining vector is \(\langle3,-2\rangle\), so each segment has length \(\sqrt{3^2+(-2)^2}=\sqrt{13}\) and the same direction.
Special cases: a point on a reflection axis stays in place; the zero translation gives zero-length joining segments. These degenerate cases do not change the central rule for distinct point-image pairs.
Check your work — Ready 1–4 answer key
- Parallel and congruent, because every point moves through the same displacement vector.
- Perpendicular bisector, because corresponding points lie at equal perpendicular distances on opposite sides of the mirror line.
- Distance, because rotation preserves the radius from the fixed center.
- They preserve distances and angle measures, so the image is congruent to the pre-image.
3. Set — Solve angles and prove congruent triangles (problems 5–10)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Missing angles (problems 5–8)
First mark what is given: triangle angle sum, linear pair, vertical angles, or parallel-line relationships. Decide whether the unknown is adjacent to, opposite from, or located at a different intersection than the known angle.
The triangle has two interior angles 23° and 85°. Its exterior angle x at the remaining vertex is the sum of the remote interior angles:
\[x=23^\circ+85^\circ=\boxed{108^\circ}.\]The adjacent interior angle is 72°; check that 72° + 108° = 180°.
The exterior angle at the right is 135° and one remote interior angle is 87°. Therefore
\[135^\circ=x+87^\circ\quad\Longrightarrow\quad\boxed{x=48^\circ}.\]Two marked lines are parallel and crossed by a transversal. The displayed 115° angle is supplementary to the corresponding acute angle. The angle x is the acute angle in the relevant matching position:
\[x=180^\circ-115^\circ=\boxed{65^\circ}.\]The two marked lines are parallel; the marked 75° angle and the angle labeled \((2x+7)^\circ\) are alternate exterior angles in the displayed arrangement, so they agree:
\[2x+7=75\quad\Longrightarrow\quad\boxed{x=34}.\]Triangle-congruence proofs (problems 9–10)
A valid SAS proof uses two corresponding sides and the angle between them. A valid AAS proof uses two corresponding angles and a side not between them. Make sure the vertex order matches the required correspondence.
Given: Y is the midpoint of \(\overline{VZ}\) and \(\overline{XW}\). Prove: \(\triangle VYW\cong\triangle ZYX\).
| Statement | Reason |
|---|---|
| Y is midpoint of \(\overline{VZ}\) and \(\overline{XW}\) | Given |
| \(\overline{VY}\cong\overline{YZ}\) and \(\overline{YW}\cong\overline{YX}\) | Definition of midpoint |
| \(\angle VYW\cong\angle ZYX\) | Vertical angles are congruent |
| \(\boxed{\triangle VYW\cong\triangle ZYX}\) | SAS |
Given: \(\angle R\cong\angle U\) and \(\overline{ST}\cong\overline{VT}\). Prove: \(\triangle SRT\cong\triangle VUT\).
| Statement | Reason |
|---|---|
| \(\angle SRT\cong\angle VUT\) | Given |
| \(\overline{ST}\cong\overline{VT}\) | Given |
| \(\angle STR\cong\angle VTU\) | Vertical angles are congruent |
| \(\boxed{\triangle SRT\cong\triangle VUT}\) | AAS (angles R ↔ U and T ↔ T; sides ST ↔ VT) |
Do not substitute SSA for AAS: the second angle at T is essential to the argument.
Check your work — Set 5–10 answer key
- \(x=108^\circ\), the sum of the two remote interior angles.
- \(x=48\), since \(x+87=135\).
- \(x=65\), since the corresponding angle adjacent to \(115^\circ\) is its supplement.
- \(x=34\), since the alternate exterior angles give \(2x+7=75\).
- Y is the midpoint of VZ and XW (given), so \(VY=YZ\) and \(WY=YX\). Also \(\angle VYW\cong\angle ZYX\) (vertical angles). Thus \(\triangle VYW\cong\triangle ZYX\) by SAS.
- \(\angle SRT\cong\angle VUT\) and \(ST=VT\) (given). The vertical angles \(\angle RTS\cong\angle UTV\); therefore the remaining angles \(\angle RST\cong\angle UVT\) by the triangle angle sum. With included side \(ST=VT\), ASA gives \(\triangle SRT\cong\triangle VUT\).
4. Go — Absolute value as a piecewise function (problems 15–18)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
This separate algebra review is part of the assigned homework. The graph of \(y=a|x-h|+k\) has vertex \((h,k)\). Each straight arm has a constant slope; for upward-opening V graphs, \(a>0\). Use the graph's vertex and a second point to determine a, then rewrite the absolute value using two cases.
At \(x=h\), either branch gives the same vertex height k. Assign the equality to just one branch to make the cases nonoverlapping.
The graph's vertex is (1,2) and both arms have slope magnitude 1. Thus
\[\boxed{y=|x-1|+2}.\]\[y=\begin{cases}-x+3,&x<1,\\x+1,&x\ge1.\end{cases}\]Check: at x=1 both expressions equal 2.
The vertex is (−3,0). In particular, y=3 at x=0, which confirms a slope magnitude of 1.
\[\boxed{y=|x+3|}\]\[y=\begin{cases}-x-3,&x<-3,\\x+3,&x\ge-3.\end{cases}\]The vertex is (−5,−4). A second point on the right arm is (−1,0), so \(a=(0+4)/(-1+5)=1\).
\[\boxed{y=|x+5|-4}\]\[y=\begin{cases}-x-9,&x<-5,\\x+1,&x\ge-5.\end{cases}\]Additional fully worked Go practice
Convert \(y=|x-4|+1\) to a piecewise function.
When \(x<4\), \(|x-4|=4-x\); when \(x\ge4\), \(|x-4|=x-4\). Therefore
\[\boxed{y=\begin{cases}-x+5,&x<4,\\x-3,&x\ge4.\end{cases}}\]An upward V graph has vertex (−2,3) and passes through (0,9). Write its absolute-value and piecewise equations.
Start with \(y=a|x+2|+3\). Substitute (0,9): \(9=2a+3\), so \(a=3\).
\[\boxed{y=3|x+2|+3}\]\[\boxed{y=\begin{cases}-3x-3,&x<-2,\\3x+9,&x\ge-2.\end{cases}}\]Check the left branch at −2: \(6-3=3\). The right branch gives \(-6+9=3\), so the branches meet at the given vertex.
Find one absolute-value expression for \(f(x)=\begin{cases}-2x-2,&x<-1,\\2x+2,&x\ge-1.\end{cases}\)
Both branches meet at (−1,0), and their slopes have magnitude 2. Thus \(\boxed{f(x)=2|x+1|}\).
Check your work — Go 15–18 answer key
The source numbering jumps from 10 to 15; no homework exercises 11–14 are printed.
- (A) \(y=|x-1|+2\). (B) \(y=\begin{cases}-x+3,&x<1\\x+1,&x\ge1.\end{cases}\)
- (A) \(y=|x|+2\). (B) \(y=\begin{cases}-x+2,&x<0\\x+2,&x\ge0.\end{cases}\)
- (A) \(y=|x+3|\). (B) \(y=\begin{cases}-x-3,&x<-3\\x+3,&x\ge-3.\end{cases}\)
- (A) \(y=|x+5|-4\). (B) \(y=\begin{cases}-x-9,&x<-5\\x+1,&x\ge-5.\end{cases}\)
5. Additional Learning Resources
Use these targeted resources after working through the explanations above. Each recommendation matches a particular part of today's assignment.
| Resource and direct link | What it teaches | Scope / What to skip |
|---|---|---|
| Khan Academy — Vertical angles practice | Identify and use vertical angles after the proof in §2. | Close match. A short reinforcement exercise. |
| Khan Academy — Proving angles are congruent (video) | Follow a geometric proof using triangle-angle and parallel-line facts. | Close match. Focus on the reasoning chain rather than the particular labels. |
| Khan Academy — Angle relationships with parallel lines | Practice identifying corresponding, alternate, and supplementary angles. | Close match. Useful for task 3–6 and Set 5–8. |
| Khan Academy — Graphing absolute value functions (video) | Read the vertex and slopes of the V-shaped graph. | Some extra material. Watch for graph-to-equation skills; solving absolute-value inequalities is not needed. |
| Khan Academy — Introduction to piecewise functions (video) | Understand how the domain is split into two rules, as in Go 15–18. | Some extra material. Concentrate on interval conditions; other kinds of piecewise functions are optional review. |
Professor Leonard was searched for a directly relevant lesson; no sufficiently close lecture was confirmed, so none is assigned. MathJax formatting requires an internet connection.
6. Teacher's review (teachers and parents)
No serious mathematical errors or omissions were identified in Section 5.6. The diagrams in main tasks 3–6 should be read together with their stated givens; apparent parallelism by itself is never a sufficient justification.
7. Section 5.6 coverage
| Packet section | Where taught |
|---|---|
| Main task 1: vertical-angle proof | In class |
| Main task 2: exterior-angle theorem proof | In class |
| Main task 3: corresponding angles | In class |
| Main task 4: alternate interior angles | In class |
| Main task 5: same-side interior angles | In class |
| Main task 6: converse alternate interior angles | In class |
| Ready 1–4: transformation features | Section 2 |
| Set 5–8: angle relationships | Section 3 |
| Set 9–10: congruence proofs | Section 3 |
| Go 15–18: absolute-value/piecewise forms | Section 4 plus three new problems |
Independent instructional companion to Mathematics Vision Project, Secondary Mathematics II, Module 5: Geometric Figures (2017), Section 5.6. Geometry and graph coordinates were mathematically constrained; graphics require no network. MathJax equation typesetting requires an internet connection.
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