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Writing a Geometric Proof: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Record only the givens

On paper, draw three distinct collinear points in the order R, S, T.

The givens are RS = 5 cm and ST = 11 cm.

The goal is to find RT.

Annotate the two given lengths and identify the whole segment sought.

Then find its length and name the geometric reason.
1. Hint 1
  The point order identifies which point is between the other two.
2. Hint 2
  Separate the two given parts from the whole requested in the goal.
Answer

Label RS as 5 cm and ST as 11 cm, with RT identified as the goal. RT = 16 cm by segment addition, since S is between R and T.

Full solution

Place S between R and T and label only the two given part lengths as supplied data.

Segment addition applies because S is between R and T, so
$RT=5+11.$
The sum is 16 centimeters.

The named points occur from left to right in the stated order. Length labels belong to the indicated parts; the lower bracket identifies the whole.

Answer

Label RS as 5 cm and ST as 11 cm, with RT identified as the goal. RT = 16 cm by segment addition, since S is between R and T.

Key idea

A diagram should distinguish information supplied from the quantity to be established.
Problem 2 Identify three properties

Name the property used in each step:

Step A: from $KL=MN$ , write $MN=KL$ .

Step B: state that segment RS is congruent to itself.

Step C: from $\angle A\cong\angle B$ and $\angle B\cong\angle C$ , conclude $\angle A\cong\angle C$ . The selected angles are unambiguous.
1. Hint 1
  One step reverses a relationship, one relates an object to itself, and one connects two relationships.
2. Hint 2
  The same reflexive, symmetric, and transitive patterns apply to congruence through equal measures.
Answer

In order: symmetric property of equality; reflexive property of segment congruence; transitive property of angle congruence.

Full solution

Step A: The first step reverses the sides of an equality, so it uses symmetry.

Step B: The segment is congruent to itself because its length equals itself, which is reflexivity.

Step C: The selected angles have equal measures in succession, so transitivity gives congruence of the first and third.

Answer

In order: symmetric property of equality; reflexive property of segment congruence; transitive property of angle congruence.

Key idea

Name a property by the change it justifies, not by the shape of the drawing.
Problem 3 Connect congruence and betweenness

Segments UV and WX are congruent, with lengths $UV=4x+3$ and $WX=47$ centimeters.

Point X lies strictly between W and Y, and $XY=24$ centimeters.

Find $x$ and $WY$ , explaining how the given congruence and point order justify the equations for the two results.
1. Hint 1
  Translate the given segment congruence into an equality of lengths.
2. Hint 2
  After finding the shared length, use the stated point order to identify the parts of WY.
Answer

x = 11; WY = 71 cm. Congruent segments have equal lengths, so the definition and substitution justify 4x + 3 = 47; X between W and Y justifies segment addition for WY.

Full solution

Congruent segments have equal lengths, so UV = WX.
Substitution gives
$4x+3=47.$

Subtract 3 from both sides to obtain
$4x=44.$

Divide both sides by the nonzero number 4 to obtain
$x=11.$

Because X is strictly between W and Y, segment addition gives
$WY=WX+XY.$

Substituting the given part lengths gives
$WY=47+24.$

Adding them gives
$WY=71\text{ cm}.$

Answer

x = 11; WY = 71 cm. Congruent segments have equal lengths, so the definition and substitution justify 4x + 3 = 47; X between W and Y justifies segment addition for WY.

Key idea

Use each geometric given for its own step: congruence supplies equal lengths, and betweenness permits adding parts.
Problem 4 Justify an angle equation

Ray OB lies inside the selected angle AOC. The selected angles AOB and BOC measure 46 and 38 degrees respectively. Find the measure of AOC and explain which hypothesis makes addition legitimate.
1. Hint 1
  Identify the two parts and the selected whole angle.
2. Hint 2
  The location of the dividing ray is the condition needed for angle addition.
Answer

The measure of AOC is 84 degrees; OB lying inside the selected angle justifies angle addition.

Full solution

The interior ray makes AOB and BOC the two parts of the selected angle AOC.

Angle addition therefore gives
$46+38=84^\circ.$

Without the interior-ray condition, the two named measures alone would not justify that sum as the selected whole.

The blue first ray and dashed red final ray bound the selected counterclockwise opening. Any named interior rays separate the consecutive parts. Use the stated angle measures and markings, rather than measuring the drawing.

Answer

The measure of AOC is 84 degrees; OB lying inside the selected angle justifies angle addition.

Key idea

An addition equation needs the geometric condition that its parts form the stated whole.
Problem 5 Supply the reasons

M is the midpoint of segment QR, whose length is 74 cm. A two-column proof has these successive statements:

Statement 1: QM = MR.

Statement 2: QM + MR = 74.

Statement 3: 2QM = 74.

Statement 4: QM = 37.

Supply a reason for each statement.
1. Hint 1
  Start by translating the midpoint definition.
2. Hint 2
  Then connect the two parts to the whole before replacing equal terms.
Answer

In order: midpoint definition and congruence-to-length equality; segment addition because M is between Q and R, then substitution of the given QR = 74; substitution and combining like terms; division of both sides by 2.

Full solution

Statement 1: The midpoint gives equal part lengths QM = MR.

Statement 2: The midpoint condition places M between Q and R, so segment addition gives QM + MR = QR.
Substituting the given whole length produces QM + MR = 74.

Statement 3: Replace MR with QM and combine the equal terms to obtain 2QM = 74.

Statement 4: Divide by the nonzero number 2 to obtain QM = 37.

Answer

In order: midpoint definition and congruence-to-length equality; segment addition because M is between Q and R, then substitution of the given QR = 74; substitution and combining like terms; division of both sides by 2.

Key idea

Every row needs a reason, including the geometric step that produces an equation.
Problem 6 Trace an angle flow proof

Ray OB lies inside the selected angle AOC and divides it into the selected parts AOB and BOC without overlapping interiors.

The selected angles $\angle BOC$ and $\angle DEF$ are congruent.

A flow proof has the following boxes:

Box I: $m\angle AOC=m\angle AOB+m\angle BOC$ .

Box II: $m\angle BOC=m\angle DEF$ .

Box III: $m\angle AOC=m\angle AOB+m\angle DEF$ .

Identify the two boxes whose arrows feed III, and supply a reason for each of the three statements.
1. Hint 1
  Which given permits an equation connecting the selected whole angle to its two parts?
2. Hint 2
  Translate the angle congruence into equality of measures before replacing a term.
Answer

The inputs are $m\angle AOC=m\angle AOB+m\angle BOC$ by angle addition and $m\angle BOC=m\angle DEF$ by congruence. Substitution gives $m\angle AOC=m\angle AOB+m\angle DEF$ .

Full solution

Box I: The interior-ray condition justifies angle addition.
With symmetry to put the whole on the left, box I states
$m\angle AOC=m\angle AOB+m\angle BOC.$

Box II: The definition of congruent angles justifies box II:
$m\angle BOC=m\angle DEF.$

Box III: Arrows from I and II feed III.
Substitution replaces the measure of BOC by the equal measure of DEF, giving
$m\angle AOC=m\angle AOB+m\angle DEF.$

Boxes I and II each have an arrow into box III. Both earlier statements are needed for the substitution in the full solution.

Answer

The inputs are $m\angle AOC=m\angle AOB+m\angle BOC$ by angle addition and $m\angle BOC=m\angle DEF$ by congruence. Substitution gives $m\angle AOC=m\angle AOB+m\angle DEF$ .

Key idea

A substitution in a flow proof depends on both an expression and an equality permitting the replacement.
Problem 7 Check the geometric conditions

For each proposed equation, state the geometric condition that must be supplied before using addition: $AB+BC=AC$ for three distinct points; $m\angle AOB+m\angle BOC=m\angle AOC$ for selected angle parts.
1. Hint 1
  One equation joins segment lengths and the other joins selected angles.
2. Hint 2
  Think about where the joining point or dividing ray must lie.
Answer

B must be between A and C. Ray OB must lie inside the selected angle AOC and divide it into the two selected parts.

Full solution

For segment addition, B must be between A and C so AB and BC form the whole AC.

For angle addition, OB must be an interior ray of the selected angle AOC, and the selected parts must be AOB and BOC on that turn.

Answer

B must be between A and C. Ray OB must lie inside the selected angle AOC and divide it into the two selected parts.

Key idea

The same algebraic shape can require different geometric hypotheses.
Problem 8 Check a plane argument

Three distinct points P, Q, and R lie in both planes $\alpha$ and $\beta$ .

An argument claims: “Three points determine exactly one plane, so these two planes must be the same.” Evaluate the argument, state the missing hypothesis that would justify its conclusion, and describe a configuration where the original givens hold but the planes are different.
1. Hint 1
  Which condition, besides distinctness, does the unique-plane foundation require?
2. Hint 2
  Test a configuration where the points share a line, and consider sheets that could meet along that line.
Answer

Noncollinearity is missing. If P, Q, and R are noncollinear, the planes coincide. The original givens also allow three distinct points on the common line of two different intersecting planes.

Full solution

The unique-plane foundation applies to three noncollinear points; distinctness alone does not supply that condition.

If noncollinearity is added, both planes contain those three noncollinear points and therefore are the unique plane through them.

For a counterconfiguration to the original claim, take two different planes intersecting along a line and choose three distinct points P, Q, and R on that line.
All three points lie in both planes even though the planes are different.

One possible configuration has P, Q and R on the intersection line of two distinct planes. Three common points need not be noncollinear.

Answer

Noncollinearity is missing. If P, Q, and R are noncollinear, the planes coincide. The original givens also allow three distinct points on the common line of two different intersecting planes.

Key idea

A uniqueness foundation applies only after all its hypotheses, including noncollinearity, are available.
Problem 9 Challenge a proposed equality

Points A, B, and C are distinct and collinear, with B between A and C. Prove by contradiction that $AB\ne AC$ . Name the accepted distance fact that the temporary assumption would violate.
1. Hint 1
  Which addition equation follows from the given point order?
2. Hint 2
  Assume the two named lengths are equal, cancel their common contribution, and compare the remaining length with its distinct endpoints.
Answer

$AB\ne AC$ ; assuming equality forces $BC=0$ , violating the positive distance between distinct points B and C.

Full solution

Assume for contradiction that $AB=AC$ .

Because B is between A and C, segment addition gives
$AC=AB+BC.$

Substitute the assumed equality to obtain
$AB=AB+BC.$

Subtract AB from both sides, giving
$BC=0.$

The distinct points B and C have positive distance, so $BC>0$ .
This contradicts $BC=0$ .
Reject the temporary equality and conclude $AB\ne AC$ .

Answer

$AB\ne AC$ ; assuming equality forces $BC=0$ , violating the positive distance between distinct points B and C.

Key idea

A contradiction can arise when an assumed equality forces distinct points to have zero distance.
Problem 10 Choose a proof route

B is between A and C, and E is between D and F.

The givens also say $AC=DF$ and $BC=EF$ .

Choose and justify a direct or contradiction approach, then write a complete paragraph proof that $AB=DE$ .

Name the conditions that permit your geometric equations.
1. Hint 1
  How do the stated point orders express the equal whole lengths as sums of parts?
2. Hint 2
  Use the equality of the removed parts to rewrite both sums with the same term before canceling it.
Answer

AB = DE. Direct cancellation suits the equal wholes and equal removed parts; B between A and C and E between D and F justify segment addition. A valid contradiction argument is also acceptable.

Full solution

Choose direct reasoning because the givens supply equal whole lengths and equal parts to remove.

Because B is between A and C, segment addition gives
$AC=AB+BC.$

Because E is between D and F, segment addition gives
$DF=DE+EF.$

Substitute these expressions into $AC=DF$ and replace EF with its equal length BC, obtaining
$AB+BC=DE+BC.$

Subtract BC from both sides to conclude
$AB=DE.$

A, B, C and D, E, F are ordered on their respective segments. The complete lengths AC and DF are given equal, and the right-hand parts BC and EF are given equal. The left-hand parts remain to be proved equal.

Answer

AB = DE. Direct cancellation suits the equal wholes and equal removed parts; B between A and C and E between D and F justify segment addition. A valid contradiction argument is also acceptable.

Key idea

Choose a route supported by the givens, and justify the geometric part equations before canceling equal terms.

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