HSG.SRT.B.4: Proving Triangle Theorems with Similarity
In plain English: HSG.SRT.B.4 is the Common Core geometry standard that asks students to prove theorems about triangles with similarity. The official examples are the triangle proportionality theorem (a line parallel to one side divides the other two sides proportionally), its converse, and a proof of the Pythagorean Theorem using similar triangles. It is usually taught in high school Geometry.
Prove theorems about triangles. Theorems include: a line parallel to one side of a triangle divides the other two proportionally, and conversely; the Pythagorean Theorem proved using triangle similarity.
Common Core State Standards for Mathematics · Domain: Similarity, Right Triangles, and Trigonometry (SRT) · Cluster: Prove theorems involving similarity Also written as HSG-SRT.B.4 or G-SRT.4 · Official standard
Students prove three theorems about triangles, each by finding a pair of similar triangles and reading off a proportion. The first is the triangle proportionality theorem, often called the side-splitter theorem: a line parallel to one side of a triangle divides the other two sides proportionally. The second is its converse: if a segment divides two sides proportionally, it is parallel to the third side. The third is the Pythagorean Theorem, proved by drawing the altitude to the hypotenuse and using the two smaller triangles it creates.
The AA criterion from HSG.SRT.A.3 does the heavy lifting in all three proofs. Students write each proof, use the theorems to find lengths and test for parallel segments, and then prove one new result, the angle bisector theorem, with the side-splitter theorem as a tool.
Learning Objectives
By the end of this lesson, students will be able to:
Prove that a line parallel to one side of a triangle divides the other two sides proportionally
Prove the converse: a segment that divides two sides of a triangle proportionally is parallel to the third side
Prove the Pythagorean Theorem using the similar triangles formed by the altitude to the hypotenuse
Use these theorems to find unknown lengths and to decide whether a segment is parallel to a side
Write a complete proof in which every statement has a reason
Prior Knowledge Required
Students should already be comfortable with:
The AA similarity criterion and its proof HSG.SRT.A.3
Angles formed by parallel lines and a transversal HSG.CO.C.9
Earlier triangle proofs with congruence HSG.CO.C.10
Explaining a proof of the Pythagorean Theorem 8.G.B.6
Hand out wide-ruled notebook paper. Its printed lines are parallel and equally spaced.
Warm-Up Prompt
"Put vertex A on one printed line. Draw sides AB and AC down to a line 6 lines below, and make the two sides different lengths. The printed lines cut AB and AC into pieces. Measure the pieces on each side. What do you notice, and does it matter how you tilted the sides?"
Students find six equal pieces on AB and six equal pieces on AC, although the pieces on the two sides have different lengths. Ask: if a printed line crosses AB two pieces down from A, how far down AC does it cross? (Also two of its six pieces.) Write the observation as AD/DB = AE/EC for one of the printed lines and tell students that today they prove it holds for every line parallel to a side, not only for notebook lines.
Direct Instruction25 minutes
Prove the three theorems in order. Use Diagram 1 for the first two and Diagram 2 for the third. After each proof, ask students to name the pair of similar triangles and the criterion that made the proof work.
Triangle proportionality (side splitter). In △ABC, D is on AB, E is on AC and DE ∥ BC. ∠ADE ≅ ∠ABC because they are corresponding angles for the parallel lines DE and BC cut by AB, and ∠A is shared. So △ADE ~ △ABC by AA, and AB/AD = AC/AE. Write AB = AD + DB and AC = AE + EC: (AD + DB)/AD = (AE + EC)/AE, so 1 + DB/AD = 1 + EC/AE. Subtract 1 and take reciprocals: AD/DB = AE/EC.
The converse. Now suppose AD/DB = AE/EC. Through D, draw the line parallel to BC (the parallel postulate gives exactly one). It meets AC at a point E′. By the side-splitter theorem, AD/DB = AE′/E′C, so AE′/E′C = AE/EC. Only one point of segment AC divides it in a given ratio, so E′ = E. The parallel line through D is line DE, so DE ∥ BC.
The Pythagorean Theorem. In right △ABC with the right angle at C, draw the altitude CD to the hypotenuse AB. △ADC and △ACB share ∠A and each has a right angle, so △ADC ~ △ACB by AA and AD/AC = AC/AB, which gives AC² = AB · AD. In the same way, △CDB ~ △ACB (shared ∠B), so BC² = AB · DB. Add: AC² + BC² = AB(AD + DB) = AB · AB = AB².
Stress two points. In the side-splitter proof, the similar triangles give AD/AB = AE/AC and DE/BC = AD/AB, but DE/BC is not equal to AD/DB. In the Pythagorean proof, the altitude must go to the hypotenuse, and AD + DB = AB is the step that ties the two proportions together. Then work these examples:
Side splitter: missing length
In Diagram 1, DE ∥ BC, AD = 6, DB = 4 and AE = 9. Find EC.
Equation: 6/4 = 9/EC, so 6 · EC = 36 and EC = 6
Converse: testing for parallel
In △ABC, AD = 10, DB = 4, AE = 12.5 and EC = 5. Is DE ∥ BC? What if EC = 5 and AE = 12 instead?
Equation: 10/4 = 2.5 and 12.5/5 = 2.5, so DE ∥ BC. With AE = 12: 12/5 = 2.4 ≠ 2.5, so DE is not parallel to BC
Pythagorean Theorem by similarity
Right △ABC has legs AC = 6 and BC = 8 and hypotenuse AB = 10 (Diagram 2). Find AD, DB and CD.
Equation: AD = AC²/AB = 36/10 = 3.6, DB = 64/10 = 6.4, and 3.6 + 6.4 = 10. △ADC ~ △CDB gives CD² = AD · DB = 23.04, so CD = 4.8
Proving a new theorem: the angle bisector theorem
In △ABC, AD bisects ∠A and meets BC at D. Draw the line through C parallel to AD; it meets line BA at F. Show BD/DC = AB/AC, then use AB = 10, AC = 6, BC = 8.
Equation: ∠AFC ≅ ∠BAD (corresponding) and ∠ACF ≅ ∠DAC (alternate interior), so △ACF is isosceles and AF = AC. Side splitter in △BCF: BD/DC = BA/AF = AB/AC. Then BD/DC = 10/6, so BD = 5 and DC = 3
Guided Practice15 minutes
Pairs work three tasks. (1) In △XYZ, M is on XY and N is on XZ with MN ∥ YZ. Write the side-splitter proof in two columns with these letters, then find NZ when XM = 5, MY = 3 and XN = 7.5. (NZ = 4.5.) (2) Test the converse on the coordinate plane: △ABC has A(0, 6), B(-3, 0), C(6, 0). D(-1, 4) is on AB and E(2, 4) is on AC. Show that AD/DB = AE/EC = 1/2 using distances, and confirm with slopes that DE ∥ BC. (3) Name the three similar right triangles in Diagram 2 with their vertices in corresponding order. Circulate and listen for pairs who write AD/DB = DE/BC; ask them to check it with the numbers in Diagram 1.
Independent Practice15 minutes
Students work four problems on their own. (1) DE ∥ BC, AD = x, DB = 5, AE = 8 and EC = 10. Find x. (x = 4.) (2) AD = 9, DB = 6, AE = 12 and EC = 8. Is DE ∥ BC? (Yes: both ratios are 3/2.) (3) The altitude to the hypotenuse of a right triangle cuts the hypotenuse into pieces of 9 and 16. Find both legs and check the Pythagorean Theorem. (The legs are 15 and 20: 15² + 20² = 625 = 25².) (4) In two or three sentences, explain why the Pythagorean proof needs the altitude to go to the hypotenuse and not to a leg.
Closure5-15 minutes
Exit ticket: (1) Write the side-splitter theorem and its converse as two if-then statements. (2) DE ∥ BC, AD = 2, DB = 5 and AE = 4. Find EC. (EC = 10.) (3) In the Pythagorean proof, which angle does each small triangle share with the large triangle? Use the longer time if students trade exit tickets and check a partner's if-then statements.
Differentiation Strategies
For Struggling Students
Give a proof frame for the side splitter with the similar triangles already named, and ask students to supply the reasons and the algebra
Have students redraw △ADE separately next to △ABC, in the same orientation, before writing any proportion
Use colored pencils: one color for the small triangle's sides, another for the pieces DB and EC, so the two kinds of ratios stay separate
For Advanced Students
Ask students to prove that three or more parallel lines cut any two transversals proportionally, using the side-splitter theorem
Ask students to prove that the altitude to the hypotenuse is the geometric mean of the two pieces of the hypotenuse
Ask students to prove the converse of the Pythagorean Theorem with a congruence argument and compare it with the similarity proof
Assessment Guidance
What to Look For
In every proof, look for the pair of similar triangles to be named with a criterion before any proportion appears. For the side splitter, check that students separate the two proportions: AD/AB = AE/AC (whole sides) and AD/DB = AE/EC (pieces). For the converse, a complete proof explains why the constructed point must be E. For the Pythagorean proof, students should state both similarities and the step AD + DB = AB.
02
Classroom Activities
3 Activities
1
Notebook-Paper Proportions
20 minPairs
Pairs use the printed lines of notebook paper as a family of parallel lines. They collect evidence for the side-splitter theorem, then test the converse by marking points that divide two sides in the same ratio.
Procedure
Each partner draws a triangle with one vertex on a printed line and the opposite side on a line 8 lines below, with different side lengths
Partners measure the pieces where the 3rd and 5th printed lines cross each side and record AD, DB, AE and EC for each line in a table
For the converse test, on a fresh triangle, draw sides that do not end on printed lines, mark D on one side and E on the other so that AD/DB = AE/EC = 2/3, and check with a ruler whether DE is parallel to the base
Then mark D and E with AD/DB = 2/3 but AE/EC = 3/2 and describe what happens
Discussion Questions
Your measured ratios were close but not exact. Why is a proof still needed?
Which step of the side-splitter proof uses the fact that the lines are parallel?
How does your converse test match the proof that uses the point E′?
Modification for Distance Learning
Use a dynamic geometry tool: construct a triangle and a line through a point on one side parallel to another side, display the four lengths and the two ratios, and drag the vertices to watch the ratios stay equal.
2
Index-Card Pythagorean Proof
25 minGroups of 3
Groups cut an index card into a right triangle, then cut along the altitude to the hypotenuse. With three triangles in hand, they match angles, write the two similarities and assemble the Pythagorean proof.
Procedure
Each group cuts two 3-by-5-inch index cards along a diagonal and keeps one right triangle from each card (two identical copies)
On one copy, fold the hypotenuse onto itself so the crease passes through the right-angle vertex; the crease is then perpendicular to the hypotenuse. Cut along the crease: this is the altitude
Label the uncut copy △ACB with the right angle at C, and the pieces △ADC and △CDB
Stack each small triangle on the uncut copy so that congruent angles line up, and color matching angles
Write the two similarity statements and the two proportions, then combine them into AC² + BC² = AB²
Discussion Questions
Which angle does each small triangle share with the uncut triangle?
Why are the two small triangles also similar to each other?
Measure your card: do the legs, 3 in and 5 in, and your measured hypotenuse fit the theorem?
Challenge Variation
Ask groups to use △ADC ~ △CDB to prove that CD² = AD · DB, and then find CD for their card from measured AD and DB.
3
Converse Detective Cards
20 minGroups of 3-4
Each group gets 6 cards. Every card shows △ABC with points D on AB and E on AC and four lengths. Groups use the converse to decide whether DE ∥ BC, then write the justification.
The 6 Cards
AD = 4, DB = 6, AE = 6, EC = 9 (parallel: both ratios 2/3)
AD = 8, DB = 2, AE = 12, EC = 4 (not parallel: 4 and 3)
AD = 3, DB = 3, AE = 11, EC = 11 (parallel: DE is a midsegment)
AD = 5, AB = 15, AE = 6, AC = 18 (parallel: DB = 10, EC = 12, both ratios 1/2)
AD = 7, DB = 5, AE = 5, EC = 7 (not parallel: 7/5 and 5/7)
AD = 2.4, DB = 3.6, AE = 3, EC = 4.5 (parallel: both ratios 2/3)
Procedure
Groups sort the cards into "parallel" and "not parallel" and write the two ratios on each card
On card 4, groups must first find DB and EC from the whole sides
Each group writes one new card of each kind and trades with another group
Discussion Questions
On card 5, the same numbers appear on both sides. Why is DE still not parallel to BC?
Is it enough to check AD/AB = AE/AC instead of AD/DB = AE/EC? Why?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: The Side-Splitter Theorem
In △ABC, drawn to scale with AB = 10 and AC = 15, DE ∥ BC. The corresponding angles at D and B and the shared angle at A make △ADE ~ △ABC, and the pieces of the two sides are in the same ratio: AD/DB = AE/EC = 3/2.
Diagram 2: The Pythagorean Theorem by Similar Triangles
Right △ABC with legs 6 and 8, drawn to scale. The altitude CD to the hypotenuse splits it into △ADC and △CDB, each similar to △ACB. The two proportions give AC² = AB · AD and BC² = AB · DB, and adding them gives AC² + BC² = AB².
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Homework Assignment
~30 min
HSG.SRT.B.4 Homework: Proving Triangle Theorems with Similarity
Directions: Show all work. In every proof, name the similar triangles and the criterion before you write a proportion, and give a reason for every statement. Leave answers as exact fractions or simplified radicals unless a problem asks for a decimal.
Part 1: The Side-Splitter Theorem and Its Converse (Problems 1-3)
In △PQR, S is on PQ and T is on PR with ST ∥ QR. (a) Prove that PS/SQ = PT/TR, giving a reason for every step. (b) If PS = 8, SQ = 6 and PT = 12, find TR.
In △JKL, M is on JK and N is on JL. (a) JM = 4, MK = 10, JN = 6 and NL = 15. Is MN ∥ KL? (b) JM = 5, MK = 7, JN = 6 and NL = 8. Is MN ∥ KL? (c) Write a proof of the converse of the side-splitter theorem: if JM/MK = JN/NL, then MN ∥ KL.
In △ABC, DE ∥ BC with D on AB and E on AC. AD = 4, DB = x, AE = 10 and EC = x + 9. Find x, DB and EC.
Part 2: The Pythagorean Theorem by Similarity (Problems 4-5)
Right △XYZ has its right angle at Z, and ZW is the altitude to the hypotenuse XY. (a) Name the two triangles similar to △XZY and give the criterion. (b) Use them to prove that XZ² + YZ² = XY². (c) If XZ = 5 and YZ = 12, find XW and WY.
The altitude to the hypotenuse of a right triangle divides the hypotenuse into segments of length 4 and 9. (a) Use similar triangles to find the length of the altitude. (b) Find both legs as simplified radicals. (c) Show that the legs and the hypotenuse satisfy the Pythagorean Theorem.
Part 3: Applying the Theorems (Problem 6)
Two straight roads meet at a point A. A fence runs from D on one road to E on the other, parallel to a third fence from B to C. AD = 45 m, DB = 30 m and AE = 54 m. (a) Find EC and name the theorem you used. (b) A farmer builds another fence from F on AB with AF = 60 m to G on AC with AG = 72 m. Use the converse to decide whether FG is parallel to BC.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Similar Triangles
Correct pair named, vertices in order, criterion stated
Correct pair without a criterion or with an order error
No similar triangles named
Proof Reasoning
Every statement has a valid reason and the logic is complete
One missing or weak reason
Several gaps or circular reasoning
Proportions and Algebra
Correct proportion and solution
Correct proportion with an arithmetic error
Wrong proportion, such as pieces compared with whole sides
Converse Decisions
Both ratios computed and the conclusion justified
Ratios correct but conclusion not justified
Decision made without ratios
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
In △ABC, DE ∥ BC with D on AB and E on AC. AD = 4, DB = 6 and AE = 5. Find EC.
Answer: C
By the side-splitter theorem, AD/DB = AE/EC, so 4/6 = 5/EC and EC = 30/4 = 7.5. Choice A uses the flipped ratio, 5 · 4/6. Choice B adds the same amount to both sides (6 - 4 = 2, so 5 + 2), which is not a proportion.
Question 2 of 20 · Multiple Choice
What is the first step in proving the side-splitter theorem for △ABC with DE ∥ BC?
Answer: B
The parallel lines make ∠ADE ≅ ∠ABC as corresponding angles, and ∠A is in both triangles, so AA gives △ADE ~ △ABC. Choice A is false: the triangles have different sizes. Choice C assumes something that is not given.
Question 3 of 20 · Multiple Choice
In △ABC, D is on AB and E is on AC. Which set of lengths shows that DE ∥ BC?
Answer: C
6/9 = 2/3 and 8/12 = 2/3, so the converse gives DE ∥ BC. Choice A gives 5/8 and 2/3. Choice B reverses the ratio on the second side (6/9 and 9/6). Choice D gives 4/5 and 5/6.
Question 4 of 20 · Multiple Choice
Right △ABC has its right angle at C, and CD is the altitude to the hypotenuse AB. Which similarity statement is correct?
Answer: D
In △ADC the right angle is at D and ∠A is shared, so A ↔ A, D ↔ C and C ↔ B: △ADC ~ △ACB. Choice A matches D with B, but ∠D is a right angle and ∠B is not. Choice C claims congruence, but the triangles have different sizes.
Question 5 of 20 · Multiple Choice
From △CDB ~ △ACB (altitude CD to the hypotenuse of right △ABC), which proportion follows?
Answer: A
Corresponding sides: CD ↔ AC, DB ↔ CB and CB ↔ AB. So DB/BC = BC/AB, which gives BC² = AB · DB. Choice B pairs DB with AB, which do not correspond: the hypotenuse AB matches the leg CB of △CDB.
Question 6 of 20 · Multiple Choice
A right triangle has legs 9 and 12 and hypotenuse 15. The altitude to the hypotenuse divides it into two segments. How long is the segment next to the leg of length 9?
Answer: B
The leg squared equals the hypotenuse times the adjacent segment: 9² = 15 · x, so x = 81/15 = 5.4. Choice A is the other segment, 144/15. Choice C is the altitude, 9 · 12/15. Choice D is half the hypotenuse.
Question 7 of 20 · Multiple Choice
DE ∥ BC in △ABC, with AD = 3, DB = x, AE = 4 and EC = x + 2. Find x.
Answer: D
3/x = 4/(x + 2), so 3(x + 2) = 4x, 3x + 6 = 4x and x = 6. Check: 3/6 = 4/8. Choice B is EC, not x. Choice A comes from setting x + 2 = 4.
Question 8 of 20 · Multiple Choice
In the proof of the converse, a student draws the line through D parallel to BC and calls its intersection with AC point E′. What must the student show next?
Answer: A
By the side-splitter theorem, AE′/E′C = AD/DB = AE/EC. Only one point of AC divides it in that ratio, so E′ = E and DE is the parallel line. Choice D is true only when D is the midpoint of AB.
Question 9 of 20 · Multiple Choice
DE ∥ BC in △ABC with D on AB and E on AC. Which proportion is NOT always true?
Answer: C
DE and BC are corresponding sides of △ADE and △ABC, so DE/BC equals AE/AC, a whole side, not AE/EC, a piece. Choices A, B and D all follow from the similarity or the side-splitter theorem.
Question 10 of 20 · Multiple Choice
DE ∥ BC in △ABC. AD = 5, DB = 10 and BC = 18. Find DE.
Answer: B
△ADE ~ △ABC, so DE/BC = AD/AB = 5/15 = 1/3 and DE = 6. Choice A uses AD/DB = 1/2. Choice C uses DB/AB, and choice D multiplies by 2 instead of dividing.
Question 11 of 20 · Multiple Choice
The altitude to the hypotenuse of a right triangle cuts the hypotenuse into segments of 3 and 12. How long is the altitude?
Answer: D
The two small triangles are similar, so 3/h = h/12, h² = 36 and h = 6. Choice A is the average of 3 and 12. Choice B divides 12 by 3.
Question 12 of 20 · Multiple Choice
In the similarity proof of the Pythagorean Theorem, the student has AC² = AB · AD and BC² = AB · DB. Why does adding them give AB²?
Answer: B
Factor out AB: AC² + BC² = AB(AD + DB), and D lies on AB, so AD + DB = AB, giving AB². Choices A and D assume an isosceles right triangle, which the theorem does not require.
Question 13 of 20 · Multiple Choice
In △ABC, AB = 12, AC = 8 and BC = 15. The bisector of ∠A meets BC at D. Find BD.
Answer: C
By the angle bisector theorem, proved with the side-splitter theorem, BD/DC = AB/AC = 12/8 = 3/2. BD = 15 · 3/5 = 9. Choice B assumes D is the midpoint of BC. Choice A uses AC/AB instead of AB/AC.
Question 14 of 20 · Multiple Choice
△ABC has A(0, 8), B(0, 0) and C(12, 0). D(0, 6) is on AB and E(3, 6) is on AC. Which statement is true?
Answer: A
AD = 2 and DB = 6, so AD/DB = 1/3. E is one quarter of the way from A to C, so AE/EC = 1/3. By the converse, DE ∥ BC; both have slope 0. Choice C is false: AD/DB would be 1 for midpoints. Choice B compares lengths on different sides.
Question 15 of 20 · Short Answer
In △FGH, J is on FG and K is on FH with JK ∥ GH. Prove that FJ/JG = FK/KH.
In △ABC, D is on AB and E is on AC. AD = 7, DB = 3.5, AE = 9 and EC = 4.5. Is DE ∥ BC? Justify with a theorem.
AD/DB = 7/3.5 = 2 and AE/EC = 9/4.5 = 2. The ratios are equal, so by the converse of the side-splitter theorem, DE ∥ BC.
Question 17 of 20 · Short Answer
A right triangle has legs a and b and hypotenuse c. The altitude to the hypotenuse divides it into segments of length x (next to leg a) and c - x (next to leg b). Use similar triangles to prove a² + b² = c².
The triangle containing leg a and segment x shares an acute angle with the whole triangle and has a right angle, so it is similar to the whole triangle by AA: x/a = a/c, so a² = cx. In the same way, the other small triangle gives (c - x)/b = b/c, so b² = c(c - x). Adding: a² + b² = cx + c² - cx = c².
Question 18 of 20 · Short Answer
A right triangle has legs 8 and 15 and hypotenuse 17. Find the two segments into which the altitude divides the hypotenuse, and check that they add to 17.
Segment next to the leg 8: 8²/17 = 64/17. Segment next to the leg 15: 15²/17 = 225/17. Check: 64/17 + 225/17 = 289/17 = 17.
Question 19 of 20 · Short Answer
A roof truss has rafters AB and AC and a horizontal tie beam DE parallel to the base BC, with D on AB and E on AC. AD = 2.4 m, DB = 1.6 m and AC = 5 m. Find AE.
DE ∥ BC, so △ADE ~ △ABC and AE/AC = AD/AB = 2.4/(2.4 + 1.6) = 0.6. AE = 0.6 · 5 = 3 m. Check with the side-splitter theorem: 2.4/1.6 = 1.5 and 3/2 = 1.5.
Question 20 of 20 · Short Answer
D is the midpoint of side AB of △ABC, and the line through D parallel to BC meets AC at E. Use the side-splitter theorem to prove that E is the midpoint of AC.
D is the midpoint, so AD = DB and AD/DB = 1. DE ∥ BC, so by the side-splitter theorem AE/EC = AD/DB = 1. Then AE = EC, so E is the midpoint of AC.
0 of 20 answered · 0 correct
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Frequently Asked Questions
10 Questions
What does HSG.SRT.B.4 mean?
HSG.SRT.B.4 means students prove theorems about triangles, here using similarity. The official list names the triangle proportionality (side-splitter) theorem, its converse, and the Pythagorean Theorem proved with similar triangles. Teachers often add results that follow from these, such as the angle bisector theorem.
Is HSG.SRT.B.4 the same as HSG.CO.C.10?
No, although both read "Prove theorems about triangles." HSG.CO.C.10 sits in the congruence domain and covers results like the angle sum and the isosceles base angles. HSG.SRT.B.4 sits in the similarity domain, so its theorems are proved with similar triangles and proportions.
What is the triangle proportionality theorem?
It says that a line parallel to one side of a triangle, crossing the other two sides, divides those sides proportionally: if DE ∥ BC, then AD/DB = AE/EC. It is also called the side-splitter theorem. The proof uses AA to show the small triangle is similar to the whole triangle.
Why prove the converse separately?
A theorem and its converse are different statements, and one being true does not make the other true. The side-splitter theorem starts from parallel lines and ends with a proportion. The converse starts from a proportion and ends with parallel lines, so it needs its own argument, usually the unique-point argument with an auxiliary parallel line.
Students learned the Pythagorean Theorem in grade 8. Why prove it again?
In grade 8 (8.G.B.6), students explain a proof, often one that rearranges areas. HSG.SRT.B.4 asks for a proof based on similar triangles, which ties the theorem to the similarity work of this unit. It also shows where the relationships a² = cx and b² = c(c - x) come from.
What mistakes do students make with the side-splitter theorem?
A frequent one is mixing pieces and whole sides, for example writing AD/DB = DE/BC. DE/BC matches AD/AB, the ratio of whole sides of the similar triangles. Other errors are flipping one ratio (AD/DB = EC/AE) and using the theorem when the segment is not known to be parallel.
Can students use SAS similarity to prove the converse?
Yes, if your course has already established SAS similarity. From AD/DB = AE/EC it follows that AD/AB = AE/AC, and with the shared angle A, SAS similarity gives △ADE ~ △ABC. Corresponding angles are then congruent, so DE ∥ BC. The auxiliary-line proof shown in this lesson avoids SAS and uses only the side-splitter theorem.
Is HSG.SRT.B.4 taught in Algebra or Geometry?
HSG.SRT.B.4 is a high school Geometry standard. It usually comes right after the AA criterion (HSG.SRT.A.3) and before right-triangle trigonometry. Students need solid proportion skills from earlier algebra courses.
What proof formats can students use?
Two-column, paragraph and flow-chart proofs are all acceptable, as long as every statement has a reason. For these theorems, a paragraph proof with a clear list of the similar triangles and the proportions is often easier to read than a long two-column proof.
Where do these theorems lead next?
The side-splitter theorem supports partitioning a segment in a given ratio on the coordinate plane (HSG.GPE.B.6). The similar right triangles of the Pythagorean proof lead to trigonometric ratios (HSG.SRT.C.6). Students also use these theorems as tools when solving problems with congruence and similarity (HSG.SRT.B.5).
07
Related Standards
6 standards
These standards connect to HSG.SRT.B.4: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSG.SRT.A.3Prerequisite
Use similarity transformations to establish the AA similarity criterion
