Last updated .
5.3 — It’s All In Your Head
In Section 5.2 you explained how a construction could justify a geometric conclusion. Here you will organize those explanations so that another reader can check every step. The essential habit is simple: every conclusion must have a reason.
1. In class: A Solidify Understanding Task
Four special lines and segments in a triangle
A triangle has several useful constructions. They can coincide in a highly symmetric triangle, but their definitions are different. Learn what each definition requires before beginning a proof.
| Object | Definition | What must be shown |
|---|---|---|
| Altitude | A segment from a vertex perpendicular to the line containing the opposite side | Starts at the vertex and forms a right angle |
| Median | A segment from a vertex to the midpoint of the opposite side | The opposite side is divided into two congruent segments |
| Angle bisector | A ray or segment from a vertex dividing its angle into two equal angles | The two smaller angles are congruent |
| Perpendicular bisector | A line perpendicular to a segment at its midpoint | Both a right angle and two congruent half-segments |
In triangle \(PQR\), point \(M\) lies on \(\overline{QR}\). Suppose \(QM=MR\) and \(PM\perp QR\). Since \(M\) is the midpoint of \(QR\), \(\overline{PM}\) is a median. Since it makes a right angle with \(QR\), it is also an altitude. The line \(PM\) is the perpendicular bisector of \(QR\).
Do these facts alone show \(PM\) bisects \(\angle QPR\)? Yes: right triangles \(PMQ\) and \(PMR\) have \(PM\) in common, \(QM=MR\), and equal included right angles. They are congruent by SAS, so \(\angle QPM\cong\angle MPR\).
Turn equality of parts into triangle congruence
Two triangles are congruent when corresponding sides and angles agree. The order of vertex names in a congruence statement tells the reader which vertices correspond: \(\triangle ABC\cong\triangle DEF\) means \(A\leftrightarrow D\), \(B\leftrightarrow E\), \(C\leftrightarrow F\).
| Congruence criterion | Information required |
|---|---|
| SSS | Three pairs of corresponding sides |
| SAS | Two pairs of corresponding sides and their included angle |
| ASA | Two pairs of corresponding angles and the included side |
After establishing triangle congruence, you may use corresponding parts of congruent triangles are congruent (often abbreviated CPCTC) to prove an angle or side equality.
Given \(JL\cong QS\), \(LM\cong SR\), and \(\angle L\cong\angle S\). The correspondence is \(J\leftrightarrow Q\), \(L\leftrightarrow S\), \(M\leftrightarrow R\). The angle \(L\) is included between \(JL\) and \(LM\), so
\[\boxed{\triangle JLM\cong\triangle QSR\quad\text{by SAS}}.\]Writing \(\triangle JLM\cong\triangle QRS\) would be wrong: it would incorrectly pair \(L\) with \(R\).
Use a flow diagram to organize reasons (main task, problems 1–8)
The packet starts with an equilateral triangle \(ABC\), and a fold that reflects the triangle onto itself through the line containing \(C\) and \(D\), where \(D\) lies on side \(AB\). A reflection fixes points on its mirror line and sends \(A\) to \(B\). Therefore that mirror line crosses \(\overline{AB}\) at its midpoint and at a right angle.
Begin with the givens. Follow arrows to results justified by definitions or known properties. When two facts are needed together (such as a midpoint and perpendicularity), show both feeding the next result.
Givens: \(\triangle ABC\) is equilateral; the line containing \(CD\) is a reflection line of the triangle, and \(D\) lies on \(AB\).
| Fact | Justification |
|---|---|
| \(AD\cong DB\) | Reflection interchanges \(A\) and \(B\) and fixes \(D\); corresponding distances are equal |
| \(D\) is the midpoint of \(AB\) | Definition of midpoint |
| \(CD\perp AB\) | A reflection axis that interchanges \(A\) and \(B\) is their perpendicular bisector |
| \(\angle ACD\cong\angle BCD\) | The reflection maps \(CA\) to \(CB\) while fixing \(CD\) |
Now apply the definitions: \(CD\) is a median because \(D\) is a midpoint; an altitude because \(CD\perp AB\); an angle bisector because the two angles at \(C\) are congruent; and the line \(CD\) is the perpendicular bisector of \(AB\) because it passes through its midpoint at right angles.
Equilateral means \(AC\cong BC\). Reflection provides \(AD\cong DB\), and \(CD\cong CD\) by the reflexive property (a segment equals itself). Consequently,
\[\triangle ACD\cong\triangle BCD\qquad(\text{SSS}).\]By corresponding parts, \(\angle ACD\cong\angle BCD\) and \(\angle ADC\cong\angle BDC\). Since \(A,D,B\) lie on a straight line, the latter two angles form a linear pair and add to \(180^\circ\); equal supplementary angles must each be \(90^\circ\). Thus \(CD\perp AB\). This is a proof using triangle congruence rather than relying only on the picture.
Convert a flow diagram to a two-column proof (main task, problem 9)
A two-column proof makes the same logical chain more linear. Put the claim in the left column and the reason beside it. Start from the givens, and never use a fact before it is established.
| Statement | Reason |
|---|---|
| 1. \(\triangle ABC\) is equilateral; \(CD\) is a line of reflection, with \(D\in\overline{AB}\). | Given |
| 2. \(AD\cong DB\) | Reflection swaps \(A\) and \(B\), fixes \(D\), and preserves distance |
| 3. \(D\) is the midpoint of \(AB\) | Definition of midpoint |
| 4. \(CD\) is a median of \(\triangle ABC\) | Definition of median |
The purpose of the table is not to make the reasoning more complicated; it lets the reader test each connection independently.
| Statement | Reason |
|---|---|
| 1. Reflection across \(CD\) interchanges \(A\) and \(B\). | Given reflection symmetry |
| 2. \(D\) is the midpoint of \(AB\). | Reflection axis between a point and its image bisects their connecting segment |
| 3. \(CD\perp AB\). | Reflection axis between a point and its image is perpendicular to their connecting segment |
| 4. The line \(CD\) is the perpendicular bisector of \(AB\). | Definition of perpendicular bisector |
For Tehani's angle-bisector claim, use the reflected rays to show \(\angle ACD\cong\angle BCD\), then cite the angle-bisector definition. For Clarita's altitude claim, use \(CD\perp AB\) and the altitude definition.
2. Ready — Write matching triangle congruence statements (homework 1–6)
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.
For the shared ideas used in these exercises, revisit the In class explanation.
First identify the common endpoint of the listed sides: it is the included-angle vertex for SAS. Then match the other two vertices using their incident sides. For SSS, use all three pairings.
Suppose \(AB\cong DE\), \(BC\cong EF\), and \(\angle B\cong\angle E\). The pairings are \(A\leftrightarrow D\), \(B\leftrightarrow E\), and \(C\leftrightarrow F\). Thus \(\boxed{\triangle ABC\cong\triangle DEF\text{ by SAS}}\).
Suppose \(PQ\cong XY\), \(QR\cong YZ\), and \(PR\cong XZ\). Then \(P\leftrightarrow X\), \(Q\leftrightarrow Y\), \(R\leftrightarrow Z\), so \(\boxed{\triangle PQR\cong\triangle XYZ\text{ by SSS}}\).
Check your work — Ready 1–6 answer key
- (a) \(\triangle MLR\cong\triangle ZJB\). (b) SAS. Plot/construction not shown; match \(M\leftrightarrow Z,L\leftrightarrow J,R\leftrightarrow B\).
- (a) \(\triangle WBP\cong\triangle QRS\). (b) SSS. Plot/construction not shown; match \(W\leftrightarrow Q,B\leftrightarrow R,P\leftrightarrow S\).
- (a) \(\triangle CYE\cong\triangle RPB\). (b) SAS. Plot/construction not shown; match \(C\leftrightarrow R,Y\leftrightarrow P,E\leftrightarrow B\).
- (a) \(\triangle CBA\cong\triangle KJM\). (b) SAS. Plot/construction not shown; match \(C\leftrightarrow K,B\leftrightarrow J,A\leftrightarrow M\).
- (a) \(\triangle DFY\cong\triangle XZW\). (b) SAS. Plot/construction not shown; match \(D\leftrightarrow X,F\leftrightarrow Z,Y\leftrightarrow W\).
- (a) \(\triangle WXZ\cong\triangle ABC\). (b) SSS. Plot/construction not shown; match \(W\leftrightarrow A,X\leftrightarrow B,Z\leftrightarrow C\).
3. Set — Identify, construct, and prove (homework 7–14)
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.
For the shared ideas used in these exercises, revisit the In class explanation.
Read the markings, not the appearance (7–10)
The Set diagram of \(\triangle GHM\) contains \(E\) on \(HM\) and \(D\) on \(GM\). The right-angle markings show \(GH\perp HM\) and \(ED\perp GM\), so \(\overline{GH}\) and \(\overline{ED}\) are altitudes of \(\triangle GHM\) (the latter from a point \(E\) on the opposite side, not a vertex of triangle \(GHM\), so it is not itself an altitude of \(\triangle GHM\)). The altitude from vertex \(H\) would have to meet line \(GM\) at a right angle. The angle arcs at \(G\) indicate \(GE\) bisects \(\angle HGM\).
For problem 10, a segment becomes a perpendicular bisector only when you mark its intersection with a side as the midpoint in addition to the shown right angle. In this diagram, mark \(GD\cong DM\), making line \(ED\) the perpendicular bisector of \(GM\) (even though \(ED\) is not an altitude of the large triangle).
Construct three different segments (11–13)
For \(\triangle DEF\), the packet asks for constructions involving side \(EF\). Because its sides are slanted, use perpendicular marks and equal-length marks rather than simply drawing what looks vertical.
| Requested construction | Procedure |
|---|---|
| Altitude from \(D\) to line \(EF\) | Through \(D\), draw a line perpendicular to the line containing \(EF\); mark its intersection \(H\). Segment \(DH\) is the altitude. Extend \(EF\) if necessary. |
| Median from \(D\) to \(EF\) | Construct midpoint \(M\) of \(EF\), and draw segment \(DM\). Mark \(EM\cong MF\). |
| Perpendicular bisector of \(EF\) | Construct the midpoint \(M\) and draw the line through \(M\) perpendicular to \(EF\). It does not need to pass through \(D\). |
Take \(D=(0,4)\), \(E=(-2,0)\), \(F=(4,0)\). The midpoint is \(M=(1,0)\), so \(DM\) is a median. But \(DM\) is slanted, while \(EF\) is horizontal: they are not perpendicular. The altitude from \(D\) is the vertical segment to \((0,0)\). The perpendicular bisector of \(EF\) is the vertical line \(x=1\).
Organize the rectangle proof (14)
In the diagram, \(ADEG\) is a rectangle, with \(A,D,C\) collinear, \(B,G,A\) collinear, and \(B,E,C\) collinear. The intended additional given is that \(\overline{ED}\) bisects \(\overline{BC}\) at \(E\), so \(BE=EC\). We can then prove \(\triangle BGE\cong\triangle EDC\).
Given: \(ADEG\) is a rectangle; \(ED\) bisects \(BC\) at \(E\); points \(A,D,C\), \(A,G,B\), and \(B,E,C\) are collinear as shown. Prove: \(\triangle BGE\cong\triangle EDC\).
| Statements | Reasons |
|---|---|
| 1. \(ADEG\) is a rectangle, and \(ED\) bisects \(BC\) at \(E\). | Given (using corrected segment \(BC\)) |
| 2. \(BE\cong EC\) | Definition of segment bisector |
| 3. \(GE\parallel AD\) and \(AG\parallel ED\) | Opposite sides of a rectangle are parallel |
| 4. \(GE\parallel DC\) and \(BG\parallel ED\) | \(DC\) lies on line \(AD\), and \(BG\) lies on line \(AG\) |
| 5. \(\angle BEG\cong\angle ECD\) | Corresponding angles formed by transversal \(BC\) cutting \(GE\parallel DC\) |
| 6. \(\angle GBE\cong\angle DEC\) | Corresponding angles formed by transversal \(BC\) cutting \(BG\parallel ED\) |
| 7. \(\triangle BGE\cong\triangle EDC\) | ASA (angle–included side–angle) |
Order check: \(B\leftrightarrow E\), \(G\leftrightarrow D\), and \(E\leftrightarrow C\). The side \(BE\) is between the two stated angles of the first triangle, just as \(EC\) is between the two stated angles of the second.
The prose and “I know” statements say \(\overline{ED}\) bisects \(\overline{BC}\), which implies \(BE\cong EC\). However, the formal “Given” line instead says it bisects \(\overline{AC}\). These are not interchangeable, and the written proof requires \(BE\cong EC\). For the intended proof, use “ED bisects BC at E.” If your teacher instructs you to use the printed \(AC\) given literally, ask for a correction; the stated ASA argument does not follow from it.
Check your work — Set 7–14 answer key
- \(\overline{GH}\) (also \(\overline{MH}\), the altitude from M).
- \(\overline{GE}\).
- \(\overline{GE}\) is not an altitude: the angle-bisector markings do not give perpendicularity, and the shown right triangle has its altitude from G along GH.
- Mark \(GD=DM\); then \(\overline{ED}\) lies on the perpendicular bisector of \(\overline{GM}\). Plot/construction not shown.
- Plot/construction not shown. Draw \(\overline{DH}\) perpendicular to line EF, with H on its extension beyond E; mark the right angle.
- Plot/construction not shown. Find midpoint N of \(\overline{EF}\), then draw \(\overline{DN}\); mark \(EN=NF\).
- Plot/construction not shown. Draw the line through midpoint N of \(\overline{EF}\), perpendicular to EF; mark the right angle and equal halves.
- A concise proof from the formal givens: \(AD=DC\) because ED bisects AC. Since ADEG is a rectangle, \(GE=AD\), so \(GE=DC\). The rectangle also gives \(\angle BGE\cong\angle EDC\) (both right angles) and \(GE\parallel DC\), so \(\angle BEG\cong\angle ECD\) by corresponding angles. Therefore \(\triangle BGE\cong\triangle EDC\) by ASA. The prose instead says ED bisects BC; using that given, \(BE=EC\) and the two corresponding-angle pairs in Tehani’s explanation also give ASA.
4. Go — Analyze transformations on a coordinate plane (homework 15–23)
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.
In the packet, the original triangle is \(ABC\) and the marked line through \(L\) and \(K\) is the mirror line for the first transformation. Later questions ask you to compare each image with the original. Use exact coordinate rules where a mirror line or translation has been specified.
| Transformation | What to do | How original–image segments behave |
|---|---|---|
| Reflect across a line (15–17) | Place each image point at the same perpendicular distance on the opposite side | The mirror line perpendicularly bisects each nonzero point-to-image segment; side lengths are preserved but orientation reverses |
| Translate right 10, down 8 (18–20) | \((x,y)\mapsto(x+10,y-8)\) | Point-to-image segments are parallel, congruent, and directed the same way; all side lengths are preserved |
| Translate down 10, then reflect over y-axis (21–23) | \((x,y)\mapsto(-x,y-10)\) | Point-to-image segments generally have different directions and lengths; corresponding triangle sides still have equal lengths |
A rigid motion preserves distances and angle measures. A reflection reverses orientation; a translation does not. A translation followed by reflection is also rigid, although the segments joining a point to its image need not be parallel.
Reflect \(P(-3,2)\), \(Q(1,5)\), and \(R(3,-1)\) across the y-axis. The rule is \((x,y)\mapsto(-x,y)\), giving \(P'(3,2)\), \(Q'(-1,5)\), \(R'(-3,-1)\). Each segment joining an original point to its image is horizontal, so it is perpendicular to the y-axis and has its midpoint on that axis. For example, the midpoint of \(P\) and \(P'\) is \((0,2)\).
Translate \(P(-3,2)\) right 10 and down 8: \(P''=(7,-6)\). The segment \(PP''\) has horizontal change \(10\), vertical change \(-8\), and slope \(\boxed{-8/10=-4/5}\). Every other original–image segment under this translation has the same slope and length \(\sqrt{10^2+(-8)^2}=\sqrt{164}=2\sqrt{41}\).
Additional fully worked Go problems
Translate triangle \(U(-2,3)\), \(V(1,-1)\), \(W(4,2)\) left 4 and up 2. List the images and compare \(UV\) to \(U'V'\).
Use \((x,y)\mapsto(x-4,y+2)\): \(U'(-6,5)\), \(V'(-3,1)\), \(W'(0,4)\). Vector \(\overrightarrow{UV}=(3,-4)\) equals \(\overrightarrow{U'V'}=(3,-4)\); the two sides are parallel and each has length \(\sqrt{3^2+(-4)^2}=5\).
Take \(X(-4,3)\), \(Y(2,-1)\), \(Z(1,5)\). Translate down 10, then reflect across the y-axis. Find the images and explain whether \(XX'\), \(YY'\), and \(ZZ'\) must be parallel.
The combined rule is \((x,y)\mapsto(-x,y-10)\), giving \(X'(4,-7)\), \(Y'(-2,-11)\), \(Z'(-1,-5)\). The joining vectors are \((8,-10)\), \((-4,-10)\), and \((-2,-10)\). They are not scalar multiples, so the joining segments are not all parallel. Yet every side length of triangle \(XYZ\) is preserved because both stages are rigid motions.
Reflect segment with endpoints \(M(-5,1)\), \(N(-1,4)\) across \(x=2\). Determine its new endpoints and verify its length is unchanged.
For reflection in \(x=2\), use \(x'=4-x\). Thus \(M'(9,1)\) and \(N'(5,4)\). Original length: \(MN=\sqrt{4^2+3^2}=5\). Image length: \(M'N'=\sqrt{(-4)^2+3^2}=5\). Hence \(\boxed{MN=M'N'}\).
In the supplied pages, the flow-diagram task is labeled problem 7 and the next tasks are 8 and 9; numbers 5 and 6 are not printed. Follow the actual numbered prompts rather than assuming missing exercises must be completed.
Check your work — Go 15–23 answer key
- Plot/construction not shown. Reflect each vertex across LK, placing \(A\prime,B\prime,C\prime\) at the same perpendicular distances on the opposite side; LK bisects each joining segment.
- The joining segments are perpendicular to LK and parallel to one another, with their midpoints on LK; they need not have equal lengths.
- Corresponding side lengths are equal. Their positions and generally their slopes differ; the triangle’s orientation reverses.
- \(A\prime\prime(2,-6),B\prime\prime(0,-3),C\prime\prime(5,-5)\). Plot not shown. Original coordinates from the labeled grid are \(A(-8,2),B(-10,5),C(-5,3)\).
- All joining segments are parallel and congruent, with vector \((10,-8)\), slope \(-\frac{4}{5}\), and length \(2\sqrt{41}\).
- Corresponding sides are congruent and parallel (or collinear); their positions change and orientation is preserved.
- \(A\prime\prime\prime(8,-8),B\prime\prime\prime(10,-5),C\prime\prime\prime(5,-7)\). Plot not shown; use \((x,y)\mapsto(-x,y-10)\).
- The joining vectors are \((16,-10),(20,-10),(10,-10)\); the segments have different slopes and lengths. Their midpoints all lie on the y-axis.
- Corresponding side lengths are equal; positions and generally slopes differ, and orientation reverses.
5. Additional learning resources
For more explanation and practice, try Khan Academy — Triangle congruence, Khan Academy — Transformations, and OpenStax — Geometry and transformations background. These are supplementary; you do not need later module techniques to complete today's packet.
6. Teacher’s review (teachers and parents)
Section 5.3, Set problem 14 — contradictory bisected segment. The paragraph and reasoning say \(ED\) bisects \(BC\), while the formal Given says it bisects \(AC\). The latter does not supply \(BE\cong EC\), required for ASA. Correction: “Given: ADEG is a rectangle; ED bisects BC at E.” Also state the collinearities shown by the diagram if requiring a fully formal proof.
Section 5.3, main task, flow-diagram labeling. Problems 5 and 6 are absent in the supplied sequence, which moves from 4 to 7. Correction: renumber consistently or explicitly note that no questions 5–6 are assigned.
7. Packet coverage
| Packet component | Instruction here |
|---|---|
| Main task 1–4: four definitions and requirements | In class and In class |
| Main task 7–8: flow diagrams, reason bank | In class |
| Main task 9: two-column proofs | In class |
| Ready 1–6: corresponding vertices, SAS and SSS | In class and Section 2 |
| Set 7–10: identify segments and markings | In class and Section 3 |
| Set 11–13: constructions | Section 3 |
| Set 14: rectangle ASA proof | Section 3, plus clarification |
| Go 15–23: reflection, translation and composition | Section 4, plus three additional worked problems |
Independent instructional companion to Mathematics Vision Project, Secondary Mathematics II: Module 5 — Geometric Figures (2017), §5.3. Not an official MVP publication. Illustrations are inline SVG; MathJax equation formatting requires an internet connection.
Last updated .