Module 02 — Transformations and Congruence
By the end you can
A few justified lines are enough when each line does a job.
Example: “I translate left 2 and up 1, so I subtract 2 from x and add 1 to y.”
P(1,−2) becomes P₁(−1,−1).
For a reflection, count equal distances from the named mirror. For a proof, name the theorem after checking its conditions.
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Write the rule or theorem, show how it applies, and state the conclusion. For a coordinate problem, show the starting, intermediate, and final coordinates. For a proof, name the evidence before the conclusion.
This review uses new rehearsal problems across the same module as Part One. It does not reproduce the real Part Two exam. Use the Explain-it page for feedback on your own words.
Triangle PQR has P(1, −2), Q(3, −2), R(1, −4). Move it left 2 and up 1, then reflect across y = 1.
Before revealing the model, write each intermediate and final coordinate. Explain why the final triangle is congruent to the original. Use the line labeled y=1 as the mirror, not the y-axis.
Compare your work one line at a time. Your wording can differ if the coordinates and reasons agree.
Given: AB ≅ DE, AC ≅ DF, and ∠BAC ≅ ∠EDF. Prove: BC ≅ EF.
Use the marked diagram. State which triangles match, name the congruence test, then justify the final side pair. Do not use CPCTC before proving the triangles congruent.
Match vertices A ↔ D, B ↔ E, C ↔ F.
ABCD is a parallelogram. Diagonal AC divides it into ΔABC and ΔCDA. Prove that opposite sides are congruent using the parallel sides, alternate interior angles, and the shared diagonal.
Then suppose AO = 7 and BO = 5. State AC and BD, and explain why they need not be equal.
Keep triangle names in corresponding order. The shared side is the included side for ASA.
Let A(0, 0), B(10, 0), C(4, 6). M is the midpoint of AC and N is the midpoint of BC. Find M and N, then show MN is parallel to AB and half as long.
The Midpoint Formula means average each coordinate: add the two x-values and divide by 2; do the same with y. No guessed positions are needed.
The midpoint between 0 and 4 is (0+4)/2=2. The midpoint between 0 and 6 is 3. Put those together: (2,3).
For the midpoint of B(10,0) and C(4,6), average 10 with 4, then 0 with 6. The result is (7,3).
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This checks the Midsegment Theorem for the stated numeric triangle. One numeric example is not a proof for every possible triangle.
For a general triangle ABC, construct LR through A parallel to BC. The angles at B and C match the two outer angles at A by alternate interior angles. Those outer angles and ∠BAC fill the straight angle LAR.
Angle Addition adds the neighboring pieces. A straight angle measures 180°. Substitute the matching angle measures to prove the triangle’s total is 180°. Do not cite Triangle Sum as its own proof.
A letter represents a number that can vary. In a general coordinate proof, we avoid picking one fixed triangle. For example, A(0,0), B(b,0), C(u,v), with b>0 and v>0, puts the base on the x-axis and C above it.
The side midpoints are (u/2,v/2) and ((b+u)/2,v/2). Their y-values match, so the midsegment is horizontal. Its length is b/2, while the base length is b. This is the same reasoning as the numeric example, now for every such triangle.
B is the name of a point. b is a number for that point’s x-coordinate. They have different jobs.
b>0 means b is positive, so B is to the right of A. v>0 puts C above the base.
The distance between u/2 and (b+u)/2 is b/2 because the two u/2 parts cancel.
Got it now? Scroll back up and re-read the section — it should land differently.
Check the requested endpoint: coordinates, a length, an angle, triangle congruence, or a particular side pair. State that endpoint in your last sentence.
Check each reason. “Given” must have been given. “Midpoint” requires equal halves on the segment. CPCTC comes after congruence. A diagram’s appearance alone does not establish a property.
These short choices check the reasons behind your written work. The Explain-it page is where you rehearse writing the responses yourself.
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