Analytic Geometry · Analytic Geometry

Multi-Step Strategies: Verifying Geometric Properties

3.3 1.1About 75 minutes

Learning goals

I can make and follow a plan to verify a geometric property in several steps.
I can verify the midsegment property of a triangle.
I can find and verify the circumcentre of a triangle by solving a linear system.
I can use variables as coordinates to prove a property for every triangle or quadrilateral.

Words to know

midsegment
A segment joining the midpoints of two sides of a triangle.

circumcentre
The point equidistant from the three vertices of a triangle; where the right bisectors meet.

multi-step strategy
A plan that breaks a problem into a sequence of smaller calculations.

proof
A logical argument that shows a statement is true in every case.

∴
The symbol for "therefore".

Putting it all together

You now have a full toolkit: midpoints, lengths, slopes, equations of lines, linear systems and circles. The last step is to use these tools together to verify a geometric property that takes several steps. That takes a plan.

This is what mathematicians, engineers and surveyors do: break a big question into small, checkable steps.

A five-part plan for any verification

Draw a diagram and label all the points.
Know / Need: write what you are given and exactly what you must show (in terms of lengths, slopes or midpoints).
Plan: list the calculations, in order.
Do: carry out each calculation and say what it shows.
Conclude: write a "Therefore..." sentence that answers the question.

Worked example

Example 1: The midsegment of a triangle

Triangle ABC has vertices A(−4, 2), B(2, 6) and C(4, −4). M is the midpoint of AB and N is the midpoint of AC. Verify that MN is parallel to BC and half its length.

Need: slope MN = slope BC, and MN = ½BC.
Plan: find M and N; find the slopes of MN and BC; find the lengths of MN and BC.
M = ((−4 + 2)/2, (2 + 6)/2) = (−1, 4). N = ((−4 + 4)/2, (2 + (−4))/2) = (0, −1).
Slope MN = (−1 − 4)/(0 − (−1)) = −5/1 = −5. Slope BC = (−4 − 6)/(4 − 2) = −10/2 = −5. Equal, so MN ∥ BC.
MN = √(1² + (−5)²) = √26. BC = √(2² + (−10)²) = √104.
Compare squares: BC² = 104 = 4 × 26 = 4 × MN². So BC = 2 × MN (since √4 = 2), that is, MN = ½BC.
Check with technology: in GeoGebra, construct the triangle and the midpoints, then drag A, B or C. The slopes stay equal and the ratio stays 1 : 2.

Therefore MN ∥ BC and MN = ½BC. The segment joining the midpoints of two sides of a triangle (a midsegment) is parallel to the third side and half as long.

M and N are midpoints (single and double tick marks show the equal halves). The dashed midsegment MN and the side BC both have slope −5, and MN is half as long as BC.

Worked example

Example 2: Finding and checking a circumcentre

Triangle ABC has vertices A(−4, 1), B(4, 5) and C(5, −2). Find the point that is equidistant from all three vertices (the circumcentre), and verify it.

Plan: the circumcentre is on the right bisector of every side. Find two right bisectors, solve the system, then check the three distances.
Right bisector of AB: midpoint (0, 3); slope AB = 4/8 = 1/2, so the perpendicular slope is −2. y = −2x + 3.
Right bisector of AC: midpoint (0.5, −0.5); slope AC = −3/9 = −1/3, so the perpendicular slope is 3. −0.5 = 3(0.5) + b, b = −2. y = 3x − 2.
Solve the system: −2x + 3 = 3x − 2, so 5 = 5x, x = 1, y = 1. Circumcentre O(1, 1).
Check: OA = √(25 + 0) = 5; OB = √(9 + 16) = 5; OC = √(16 + 9) = 5 ✓.

Therefore O(1, 1) is equidistant from A, B and C (5 units). A circle with centre (1, 1) and radius 5 passes through all three vertices.

The two right bisectors y = −2x + 3 and y = 3x − 2 meet at the circumcentre O(1, 1). The dashed circle with centre O and radius 5 passes through all three vertices.

Worked example

Example 3: Square diagonals

Verify that S(1, 2), T(5, 4), U(3, 8), V(−1, 6) is a square and that its diagonals are perpendicular bisectors of each other.

Square: ST = √(16 + 4) = √20; TU = √(4 + 16) = √20; UV = √20; VS = √20. Slopes ST = 2/4 = 1/2 and TU = 4/(−2) = −2: product −1. Four equal sides and a right angle: a square.
Diagonals: SU from (1, 2) to (3, 8): midpoint (2, 5), slope 6/2 = 3. TV from (5, 4) to (−1, 6): midpoint (2, 5), slope 2/(−6) = −1/3.
Same midpoint, so they bisect each other. 3 × (−1/3) = −1, so they are perpendicular.

Therefore STUV is a square, and its diagonals are perpendicular bisectors of each other (and equal: SU = TV = √40).

Worked example

Example 4: Points on a circle

Verify that the triangle with vertices P(3, 4), Q(−5, 0) and R(4, −3) has all three vertices on the circle x² + y² = 25, and use this to name its circumcentre.

P: 9 + 16 = 25 ✓. Q: 25 + 0 = 25 ✓. R: 16 + 9 = 25 ✓.
All three are on the circle with centre (0, 0) and radius 5, so each is 5 units from the origin.

Therefore the circumcentre of triangle PQR is the origin (0, 0).

Going further: a proof for every triangle

Example 1 verified the midsegment property for one triangle. To prove it for every triangle, use variables for the coordinates. Choose them cleverly to keep the algebra simple: put one vertex at the origin and one side on the x-axis, and use 2a, 2b, 2c so the midpoints have no fractions.

Worked example

Example 5: The midsegment property in general

Prove that in any triangle, the segment joining the midpoints of two sides is parallel to the third side and half its length.

Place the triangle with A(0, 0), C(2a, 0) on the x-axis (a > 0), and B(2b, 2c) anywhere above it (c ≠ 0). Every triangle can be placed this way.
Midpoint of AB: M(b, c). Midpoint of BC: N((2b + 2a)/2, (2c + 0)/2) = N(a + b, c).
MN goes from y = c to y = c: slope 0. AC goes from y = 0 to y = 0: slope 0. So MN ∥ AC.
MN = (a + b) − b = a. AC = 2a − 0 = 2a. So MN = ½AC.

Therefore, for every triangle, the midsegment is parallel to the third side and half as long. A proof with variables covers all cases at once; a numerical check covers only one.

A general triangle placed with A at the origin and AC on the x-axis. Because both midpoints have y-coordinate c, the dashed midsegment MN is horizontal like AC, and its length a is half of 2a.

Just for fun

Mathematicians love the word "therefore" so much that it has its own symbol: ∴ (three dots in a triangle). It saves ink, and it looks like a tiny stack of oranges at the grocery store.

Hands-on activity

Design and verify your own property

You need: graph paper (or GeoGebra on a computer), a ruler, a pencil and a calculator.

Pick one property to test from this list: (a) the diagonals of an isosceles trapezoid are equal; (b) joining the midpoints of the sides of a rectangle makes a rhombus; (c) the midpoint of the hypotenuse of a right triangle is the same distance from all three vertices.
Draw an example on graph paper with whole-number coordinates. (Try to make it lopsided, not a special case.)
Write a Know / Need / Plan for your property.
Carry out the plan with the formulas. Write a concluding sentence.
Test a second, different example. Does the property still hold?
Explain to a grown-up why two examples are convincing but not a proof, and how a proof with variables (like Example 5) fixes that.

What to look for: a clear plan before calculating, correct use of formulas, and a concluding sentence. Sample for (c): right triangle (0, 0), (8, 0), (0, 6); midpoint of hypotenuse (4, 3); distances to all three vertices are √(16 + 9) = 5 ✓.

Your turn

Triangle PQR has P(0, 0), Q(6, 8), R(10, 0). D is the midpoint of PQ and E is the midpoint of QR. Verify that DE ∥ PR and DE = ½PR.

Find the circumcentre of the triangle with vertices (0, 0), (6, 0) and (0, 8), and verify it.

Verify that the diagonals of the rectangle (0, 0), (8, 0), (8, 6), (0, 6) bisect each other and are equal.

Verify that the triangle with vertices (6, 8), (−10, 0) and (0, −10) has its circumcentre at the origin.

Write a Know / Need / Plan for: "Verify that the quadrilateral A(−2, 0), B(2, −2), C(4, 2), D(0, 4) is a square." Then carry it out.

More practice: warm-up

What is a midsegment of a triangle?

What two facts does the midsegment property state?

The circumcentre is where the ___ of the sides meet.

What symbol means "therefore"?

Why do we use 2a, 2b, 2c in general proofs about midpoints?

More practice: on your own

Triangle ABC has A(−6, −2), B(2, 6), C(6, −4). Verify that the midsegment joining the midpoints of AB and BC is parallel to AC and half its length.

Find the circumcentre of the triangle (−2, 0), (4, 0), (2, 4) and its distance from each vertex.

Verify that the diagonals of the rhombus (0, 0), (4, 3), (9, 3), (5, 0) bisect each other.

Verify that the midpoint of the hypotenuse of the right triangle (−2, −1), (4, −1), (4, 7) is equidistant from all three vertices.

Verify that A(1, 1), B(7, 3), C(5, 7), D(−1, 5) is a parallelogram in two different ways.

Challenge

Use A(0, 0), B(2a, 0), C(2a, 2b), D(0, 2b) to prove that the diagonals of every rectangle are equal and bisect each other.

Use A(0, 0), B(2a, 0), C(0, 2b) to prove that the midpoint of the hypotenuse of a right triangle is equidistant from all three vertices.

Prove Varignon's theorem for a special case: the quadrilateral A(0, 0), B(2a, 0), C(2b, 2c), D(2d, 2e). Show the midpoint quadrilateral has opposite sides with equal slopes.

Check your understanding

Quiz

A midsegment of a triangle is:

To find the circumcentre analytically, you should:

What is the first step of a good verification plan?

M(−1, 4) and N(0, −1) are midpoints of AB and AC, and BC = √104. What is MN?

Why does a proof with coordinates such as A(0, 0), B(2b, 2c), C(2a, 0) cover every triangle?

A triangle's vertices all satisfy x² + y² = 49. Its circumcentre is:

For parents and teachers

Big idea: multi-step verifications combine midpoints, lengths, slopes, equations of lines and linear systems, guided by a written plan; using variables turns a single check into a proof for every case. This lesson teaches expectation 3.3 (plan and implement a multi-step strategy that uses analytic geometry and algebraic techniques to verify a geometric property). Example 1 is the curriculum's own example, and Example 2 combines it with solving linear systems (expectation 1.1).

How to help: Ask your teen to read their plan aloud before calculating, and to finish with "Therefore...". The general proofs (Example 5 and the challenge questions) are a great stretch for strong students and a preview of senior high school mathematics.

Common difficulties:

Starting calculations without a plan and losing track of what is needed.
Arithmetic slips in long problems: encourage checking each step (e.g. checking that a found circumcentre is the same distance from all three vertices).
Comparing lengths with square roots: compare the squares instead (BC² = 4 × MN² means BC = 2MN).
Thinking one numerical example proves a general rule.