Last updated .
5.2 — Do You See What I See?
In Section 5.1, rotating a triangle helped explain why its interior angles sum to 180°. This time you will often see a finished diagram and must explain which steps and known properties make its patterns trustworthy. The goal is a clear mathematical explanation, not yet a formal two-column proof.
1. In class: A Develop Understanding Task
What does a diagram really tell you?
A diagram can help you formulate a conjecture—a claim that appears true. But lengths and angles that merely look equal are not established facts. You can rely on a stated given, construction steps, explicitly marked congruent segments or angles, and geometric results you already know.
Identify what is given → state the claim → connect the two using definitions, established facts, and justified construction steps → explain why the conclusion follows.
A sketch shows a triangle with two sides that appear the same length. May we immediately conclude that it is isosceles?
No. The drawing alone is not a measurement or a given. If the two sides are explicitly marked congruent, or both are radii of the same circle, then we can justify their equality. Only then may we call the triangle isosceles.
Reconstruct the story behind intersecting circles (main task, problems 1–3)
Start with points \(A\) and \(B\). Draw one circle centered at \(A\) passing through \(B\), and another centered at \(B\) passing through \(A\). Let \(C\) be an intersection of the circles.
Because \(C\) is on the circle centered at \(A\), \(AC=AB\). Because \(C\) is on the circle centered at \(B\), \(BC=BA\). Therefore \(AC=AB=BC\). All three sides are equal, so \(\triangle ABC\) is equilateral; consequently, its three angles each measure \(60^\circ\).
Construction sequence: draw \(\overline{AB}\), draw the circles with radius \(AB\), mark their intersection \(C\), and connect \(A\), \(B\), and \(C\). Notice that the equality comes from the circle definitions—not from how the printed picture looks.
Reason about the quadrilateral formed by two circles (main task, problems 4–6)
Now consider congruent circles centered at \(A\) and \(B\), intersecting at \(C\) (above) and \(D\) (below). In the packet, \(P\) is the intersection of the diagonals \(\overline{AB}\) and \(\overline{CD}\). The packet explicitly gives that the two circles are congruent.
All four segments \(AC,BC,AD,BD\) are radii of two congruent circles, so \(AC=BC=AD=BD\). Hence quadrilateral \(ACBD\) has four equal sides: it is a rhombus, which is also a parallelogram. Its opposite sides are parallel.
Since \(C\) and \(D\) each have equal distance from \(A\) and \(B\), both lie on the perpendicular bisector of \(AB\). Thus \(CD\perp AB\) and the diagonals cross at the midpoint \(P\) of \(AB\). Symmetry also shows that \(P\) is the midpoint of \(CD\).
As an alternative proof of one diagonal property: triangles \(ACB\) and \(ADB\) are congruent by SSS (\(AC=AD\), \(BC=BD\), and common side \(AB\)); reflection across \(AB\) interchanges \(C\) and \(D\).
Claim: \(ACBD\) is a rhombus.
“The circles centered at \(A\) and \(B\) are congruent, so their radii have equal length. Each of the segments \(AC\), \(AD\), \(BC\), and \(BD\) is a radius of one of the circles. Therefore all four sides of \(ACBD\) are congruent. A quadrilateral with four congruent sides is a rhombus.”
This paragraph contains the given, the relevant fact, and the conclusion in a readable order. For a second conjecture, you could justify why the diagonals are perpendicular.
Why does a perpendicular bisector create equal distances? (main task, problems 7–10)
The midpoint \(M\) of \(\overline{AB}\) satisfies \(AM=MB\). The perpendicular bisector of \(\overline{AB}\) is the line through \(M\) that forms right angles with \(AB\). Any point \(C\) on this line is called equidistant from \(A\) and \(B\) when \(CA=CB\).
Given: \(M\) is the midpoint of \(\overline{AB}\), line \(\ell\) is perpendicular to \(AB\) at \(M\), and \(C\) is any point on \(\ell\).
Goal: prove \(AC=BC\).
1. Midpoint definition gives \(AM=MB\).
2. Perpendicular lines make right angles, so \(\angle AMC\cong\angle BMC\) (both measure \(90^\circ\)).
3. Segment \(CM\) is shared by both triangles, so \(CM=CM\).
4. The triangles have side–included angle–side congruent; therefore \(\triangle AMC\cong\triangle BMC\) by SAS.
5. Corresponding sides of congruent triangles are congruent. Therefore \(AC=BC\): point \(C\) is equidistant from the endpoints.
We did not choose a special position for \(C\). The same argument works above or below \(AB\), and at \(C=M\) the equality follows immediately from \(AM=BM\).
Do not reverse the direction of this statement without giving a separate justification; today the task is to establish the forward direction.
How to record the reasoning with symbols
| Readable statement | Mathematical notation |
|---|---|
| M is the midpoint of AB | \(AM\cong MB\) |
| The bisector is perpendicular to AB | \(\ell\perp\overleftrightarrow{AB}\) |
| Two included angles are right angles | \(\angle AMC\cong\angle BMC\) |
| The triangles share a side | \(\overline{CM}\cong\overline{CM}\) |
| The two triangles are congruent | \(\triangle AMC\cong\triangle BMC\) |
| Conclusion | \(AC=BC\) |
In packet problems 8–9, write these statements in their logical order, with the goal clearly identified. The word any in problem 10 matters: the point \(C\) was arbitrary, so the argument establishes a general property.
2. Ready — geometry symbols (homework problems 1–14)
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.
Symbols let you distinguish a geometric object from its measure. In particular, \(\overline{GH}\) is a segment, whereas \(GH\) without a bar often denotes its length; be alert to fonts and notation in printed work.
| Symbol | Meaning | Example / reminder |
|---|---|---|
| \(=\) | Equal numerical values | \(GH=8\) |
| \(m\angle C\) | Measure of angle C | \(m\angle C=60^\circ\) |
| \(\overline{GH}\) | Segment with endpoints G and H | Finite geometric object |
| \(\triangle ABC\) | Triangle ABC | Vertices named in order |
| \(\perp\) | Perpendicular | Forms a right angle |
| \(\angle ABC\) | Angle with vertex B | Middle letter = vertex |
| \(\overleftrightarrow{GH}\) | Line through G and H | Extends in both directions |
| \(\cong\) | Congruent | Same size and shape |
| \(\sim\) | Similar | Same shape, not necessarily size |
| \(GH\) | Length of segment GH | A numerical distance |
| \(\overrightarrow{GH}\) | Ray starting at G through H | First letter = endpoint |
| \(\parallel\) | Parallel | Lines in a plane that do not meet |
| \(\pm\) | Plus or minus | \(x=\pm 3\) |
| \(|x|\) | Absolute value | Distance from zero, never negative |
Check your work — Ready 1–14 answer key
- F — equal values.
- J — angle measure.
- N — segment length.
- H — triangle ABC.
- K — perpendicular.
- L — angle ABC.
- P — line GH.
- B — congruent.
- M — similar.
- D — segment GH.
- E — ray GH (endpoint G).
- C — parallel.
- G — plus or minus.
- A — absolute value, distance from zero. The definition should say nonnegative, since \( |0|=0\).
3. Set — compass constructions and angle algebra (homework problems 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.
Construct a perpendicular bisector (15–16)
With a compass, use the same opening (greater than half the segment length) to draw arcs centered at each endpoint, one arc above the segment and one below. The two pairs of arcs intersect at points equidistant from both endpoints. Connect the two intersection points with a straightedge. This is the perpendicular bisector; mark the right angle with a small square and mark the two halves of the original segment with matching tick marks.
Given \(\overline{PQ}\), construct line \(\ell\) by the equal-radius arc method. If \(M=\ell\cap PQ\), then \(PM=MQ\) and \(\ell\perp PQ\). These are two separate facts to mark: congruent halves and the right angle.
Construct an angle bisector (17–19)
An angle bisector is a ray that divides an angle into two congruent angles. Put the compass point at the vertex and draw an arc cutting both sides at points \(P\) and \(Q\). Keeping the same compass width, draw intersecting arcs centered at \(P\) and \(Q\) inside the angle. Draw the ray from the original vertex through that intersection. Mark the two equal angles with matching arcs.
Ray \(CF\) bisects \(\angle ECD\), and \(m\angle ECF=2x+10\), \(m\angle FCD=3x-18\). Find \(m\angle FCE\).
\[2x+10=3x-18\]\[x=28\]\[m\angle FCE=2(28)+10=\boxed{66^\circ}.\]The requested angle is \(FCE\), not the whole \(ECD\); the whole angle is \(132^\circ\).
Use stated right angles to find missing measures (20–22)
For a right angle split into parts, the parts add to \(90^\circ\). For a triangle, all three interior angles add to \(180^\circ\). For a right triangle, its other two angles therefore add to \(90^\circ\). Use the figure’s labels to build an equation; do not assume a right angle solely from the way a segment is drawn.
A right triangle has acute angles \((3x+2)^\circ\) and \((2x+8)^\circ\). Find \(x\) and both angle measures.
\[(3x+2)+(2x+8)=90\]\[5x+10=90\Longrightarrow x=16.\]Then the angles are \(50^\circ\) and \(40^\circ\). Check that \(90+50+40=180\).
Check your work — Set 15–23 answer key
- Plot/construction not shown. Construct the perpendicular bisector of \(\overline{AB}\); its intersection M satisfies \(AM=MB\). Mark the right angle and equal half-segments.
- Plot/construction not shown. Construct the perpendicular bisector of \(\overline{WV}\); mark its right angle and equal half-segments.
- Plot/construction not shown. Draw a ray from B splitting \(\angle ABC\) into two congruent angles; mark matching arcs.
- Plot/construction not shown. Draw a ray from E splitting \(\angle DEF\) into two congruent angles; mark matching arcs.
- Plot/construction not shown. Draw a ray from H splitting \(\angle GHI\) into two congruent angles; mark matching arcs.
- \(x=24\text{ cm}\), since \(x^2=26^2-10^2=576\).
- \(x=30^\circ\), since \(x+90^\circ+2x=180^\circ\).
- \(x=\sqrt{485-136\sqrt{10}}\text{ m}\approx7.411\text{ m}\). Pythagoras gives \(BD=17\), \(BE=\sqrt{14^2-6^2}=4\sqrt{10}\), and \(ED=17-4\sqrt{10}\); then \(x^2=ED^2+6^2\).
- \(x=28\) and \(m\angle FCE=66^\circ\). The bisector gives \(2x+10=3x-18\); substitute into \(2x+10\).
4. Go — translations, reflections, and rotations (homework problems 24–28)
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.
These transformations were introduced earlier. Today you apply coordinate rules and notice invariant properties: translations preserve slope of corresponding-point connections, reflections cross their mirror line perpendicularly, and half-turns preserve distances.
| Transformation | Point rule | What stays true |
|---|---|---|
| Right 6 and down 4 | \((x,y)\mapsto(x+6,y-4)\) | Every point moves by the same vector |
| Reflect across \(x=0\) (y-axis) | \((x,y)\mapsto(-x,y)\) | Connecting segments to images are horizontal |
| Rotate \(180^\circ\) about \((-2,0)\) | \((x,y)\mapsto(-4-x,-y)\) | Center is midpoint of original/image pair |
Original points: \(C(-4,0)\), \(E(-3,3)\), \(D(0,4)\).
| Transformation | C image | E image | D image |
|---|---|---|---|
| Right 6, down 4 (primed) | \(C'(2,-4)\) | \(E'(3,-1)\) | \(D'(6,0)\) |
| Reflect across \(x=0\) (double-primed) | \(C''(4,0)\) | \(E''(3,3)\) | \(D''(0,4)\) |
| Rotate 180° about \((-2,0)\) (triple-primed) | \(C′′′(0,0)\) | \(E′′′(-1,-3)\) | \(D′′′(-4,-4)\) |
For the translation, the slopes of \(\overline{CC'}\), \(\overline{EE'}\), and \(\overline{DD'}\) are all \(\boxed{-4/6=-2/3}\). For the reflection, \(D\) lies on the mirror line, so \(D''=D\) and no distinct connecting segment is needed. The segments \(\overline{CC''}\) and \(\overline{EE''}\) are perpendicular to \(x=0\) and bisected by it.
Additional fully worked Go problems
Translate triangle with vertices \(P(-2,1)\), \(Q(1,4)\), \(R(3,-1)\) left 3 and up 2. Determine the images and slope of \(\overline{PP'}\).
The rule is \((x,y)\mapsto(x-3,y+2)\). Thus \(P'(-5,3)\), \(Q'(-2,6)\), and \(R'(0,1)\). Every connecting segment has slope \(\Delta y/\Delta x=2/(-3)=\boxed{-2/3}\).
Rotate triangle \(J(1,-2)\), \(K(-4,3)\), \(L(0,5)\) through 180° about \((2,1)\). Find the image and check one midpoint.
The half-turn rule about \((h,k)\) is \((x,y)\mapsto(2h-x,2k-y)\). Here it becomes \((4-x,2-y)\). Therefore \(J'(3,4)\), \(K'(8,-1)\), and \(L'(4,-3)\).
The midpoint of \(J(1,-2)\) and \(J'(3,4)\) is \(\bigl((1+3)/2,(-2+4)/2\bigr)=(2,1)\), exactly the given rotation center.
Reflect \(T(-1,4)\) and \(U(5,-2)\) over \(x=2\), and explain the relationship of each point-to-image segment to the mirror line.
The reflected x-coordinate is \(x'=2(2)-x=4-x\); the y-coordinate stays the same. Therefore \(T'(5,4)\) and \(U'(-1,-2)\). Segments \(TT'\) and \(UU'\) are horizontal (perpendicular to the vertical line \(x=2\)), and each is bisected by the mirror line.
The printed diagram says the circles centered at A and B are congruent. That is why radii from both circles can be treated as equal, supporting the rhombus and perpendicular-diagonal conclusions. Mere visual similarity of the circles would not be enough.
The construction shows a particular point C in the picture, but the SAS argument applies to any C on the perpendicular bisector. The special case C=M is handled directly by the midpoint property rather than by a nondegenerate triangle congruence argument.
Use the original C, E, D positions for the double-primed reflection, not the already translated C′, E′, D′ points. Similarly the triple-primed 180° rotation in problem 28 starts from the original triangle. The point D lies on the reflection line, so its image coincides with D.
Check your work — Go 24–28 answer key
- Assuming one grid square represents one unit: \(C(-4,0),E(-3,3),D(0,4)\). The source graph does not label a numerical scale, so these coordinates use that assumption.
- \(C\prime(2,-4),E\prime(3,-1),D\prime(6,0)\), using the one-unit grid assumption from 24. Plot not shown.
- Each slope is \(-\frac{4}{6}=-\frac{2}{3}\). Plot not shown; all three joining segments have displacement \((6,-4)\).
- \(C\prime\prime(4,0),E\prime\prime(3,3),D\prime\prime(0,4)\), using the grid assumption from 24. Plot not shown. D is on the mirror line and stays fixed; the other joining segments are perpendicular to \(x=0\) and bisected by it.
- \(C\prime\prime\prime(0,0),E\prime\prime\prime(-1,-3),D\prime\prime\prime(-4,-4)\), using the grid assumption from 24 and rule \((x,y)\mapsto(-4-x,-y)\). Plot not shown.
5. Teacher’s review (teachers and parents)
Section 5.2, main task, problem 7, statement immediately following: The perpendicular-bisector theorem is written with a question mark despite being the claim students are to prove. Suggested correction: State it declaratively: “Every point on the perpendicular bisector of a segment is equidistant from the segment’s endpoints.”
Section 5.2, main task, problems 8–10: SAS with triangles AMC and BMC assumes C is not the midpoint M. If C=M the triangles degenerate, although the conclusion remains true. Suggested correction: State that C is on the perpendicular bisector and C≠M for the SAS step, then handle C=M directly using AM=BM.
Section 5.2, Go, problems 27–28: The notation distinguishes successive image labels but does not explicitly say each transformation is performed on the original diagram. Suggested correction: Specify “Return to the original triangle CED before each transformation” if that is the intended independent practice.
6. Packet coverage
| Packet portion | Where it is taught |
|---|---|
| Main task 1–3: intersecting circles and triangle | In class |
| Main task 4–6: circle quadrilateral and paragraph arguments | In class |
| Main task 7–10: perpendicular bisector, symbolic reasoning | In class |
| Ready 1–14: geometry symbols | Section 2 |
| Set 15–19: compass constructions | Section 3 |
| Set 20–23: right angles and angle bisectors | Section 3 |
| Go 24–28: coordinate transformations | Section 4 plus three original worked problems |
Independent instructional companion to Mathematics Vision Project, Secondary Mathematics II: Module 5 — Geometric Figures (2017), §5.2. Not an official MVP publication. Text and SVG diagrams are embedded; MathJax equations require an internet connection.
Last updated .