Prove that a line parallel to one side of a triangle divides the other two proportionally, prove the converse, and prove the Pythagorean Theorem using similar triangles
Several sessions: ruler-and-protractor experiments or coordinate work, a proof discussed as a class, modeled examples, independent practice and a written argument.
Ohio learning standards this draft targets
math-2017:G.SRT.4
Read the standard text
math-2017:G.SRT.4 Prove and apply theorems about triangles. Theorems include but are not restricted to the following: a line parallel to one side of a triangle divides the other two proportionally, and conversely; the Pythagorean Theorem proved using triangle similarity.
Provisional targets. Clause-by-clause educational review is still pending.
📘Section 1 · LearnLearn the idea
Right triangle ABC with the right angle at C, legs b = AC = 6 and a = BC = 8, and hypotenuse c = AB = 10. The altitude CD = 4.8 meets the hypotenuse at D, splitting it into AD = 3.6 and BD = 6.4. Triangle ACD is similar to ABC, giving 6² = 10 × 3.6, and triangle CBD is similar to ABC, giving 8² = 10 × 6.4; adding gives 36 + 64 = 100 = 10². The drawing is 40 pixels per unit.
Learning goal: Prove and apply theorems about triangles, including these: a line parallel to one side of a triangle divides the other two sides proportionally, and conversely; and the Pythagorean Theorem proved using triangle similarity.
Tools available. The proofs use the AA criterion for similar triangles, the angle facts for parallel lines cut by a transversal (corresponding angles are equal), and the fact that similar triangles have corresponding sides in one ratio. Each proof is written so that every step names its reason.
Theorem 1: a line parallel to one side divides the other two proportionally. Let triangle ABC have a point D on side AB and a point E on side AC, with line DE parallel to BC. Claim: AD/DB = AE/EC. Proof. Line AB is a transversal of the parallel lines DE and BC, so angle ADE equals angle ABC, corresponding angles. Angle DAE is the same angle as angle BAC; both triangles have it at A. Two pairs of equal angles, so triangle ADE is similar to triangle ABC by AA, with D ↔ B and E ↔ C. Therefore AD/AB = AE/AC. Now the algebra that a student would skip: take reciprocals to get AB/AD = AC/AE, write AB = AD + DB and AC = AE + EC, so (AD + DB)/AD = (AE + EC)/AE, which is 1 + DB/AD = 1 + EC/AE, so DB/AD = EC/AE, and taking reciprocals once more, AD/DB = AE/EC. The same similarity also gives DE/BC = AD/AB, so the parallel segment is the same fraction of BC that AD is of AB.
A worked model: AB = 12, AC = 9, and D is on AB with AD = 8, so DB = 4. If DE is parallel to BC, then AE/AC = AD/AB = 8/12 = 2/3, so AE = 6 and EC = 3. Check the theorem's statement: AD/DB = 8/4 = 2 and AE/EC = 6/3 = 2. If BC = 10.5, then DE = (2/3)(10.5) = 7.
Theorem 2, the converse: if a line divides two sides proportionally, it is parallel to the third side. Let D be on AB and E on AC with AD/DB = AE/EC. Claim: DE is parallel to BC. Proof. Draw the line through D parallel to BC. It is not parallel to line AC (otherwise AC would be parallel to BC, but they meet at C), and it enters the triangle at D and cannot leave through side BC, which it never meets, so it crosses side AC at a point between A and C; call it E*. By Theorem 1, AD/DB = AE*/E*C. But we were given AD/DB = AE/EC. So AE*/E*C = AE/EC. A point on segment AC that divides it in a given ratio is unique: if the point is x from A, the ratio is x/(AC − x), which increases strictly as x increases from 0 toward AC (the numerator grows and the denominator shrinks), so it takes each positive value exactly once. So E* = E, and the line DE is the line through D parallel to BC. This style of proof, constructing the object you want and showing it coincides with the given one, is common in geometry. It does not need any similarity criterion beyond the one already used.
Application: D on AB with AD = 5, DB = 7.5, and E on AC with AE = 4, EC = 6. Then AD/DB = 5/7.5 = 2/3 and AE/EC = 4/6 = 2/3, so DE is parallel to BC by the converse. If instead EC were 7, the ratios 2/3 and 4/7 would differ and DE would not be parallel to BC.
Theorem 3: the Pythagorean Theorem, proved using similarity. Let triangle ABC have a right angle at C, with legs a = BC and b = AC and hypotenuse c = AB. Drop the altitude from C to the hypotenuse, meeting AB at D. D lies between A and B, because angles A and B are both acute (they add to 90°), so the foot of the perpendicular from C cannot fall outside segment AB. Claim: a² + b² = c². Proof. Triangle ACD and triangle ABC share the angle at A, and each has a right angle, at D and at C, so they are similar by AA, with the correspondence A ↔ A, C ↔ B, D ↔ C. Corresponding sides give AC/AB = AD/AC, so AC² = AB · AD, that is, b² = c · AD. Likewise triangle CBD and triangle ABC share the angle at B and each has a right angle, so they are similar by AA, giving BC/AB = BD/BC, so a² = c · BD. Add: a² + b² = c · AD + c · BD = c(AD + BD) = c · AB = c². The altitude split the hypotenuse into two pieces, and each leg squared equals the whole hypotenuse times the piece next to that leg.
A worked model: legs 6 and 8, hypotenuse 10, with a = BC = 8 next to B and b = AC = 6 next to A. Then AD = b²/c = 36/10 = 3.6 and BD = a²/c = 64/10 = 6.4, and 3.6 + 6.4 = 10, the whole hypotenuse. The altitude is CD = ab/c = 48/10 = 4.8, because the area is ½ · 6 · 8 = 24 and also ½ · 10 · CD. The two small triangles are also similar to each other, which gives CD² = AD · BD: 4.8² = 23.04 and 3.6 · 6.4 = 23.04. Every number agrees.
Why these are theorems and not observations. A single drawing with AD = 8, AE = 6 shows the ratio holding once. The proof shows it holds for every triangle and every parallel line, because it used only the angle facts for parallel lines and the AA criterion, neither of which depends on the particular lengths. The converse is a separate statement with a separate proof; students often assume that proving a theorem proves its converse, and it does not. "If DE is parallel then the ratios are equal" and "if the ratios are equal then DE is parallel" are different claims, and the second needed the uniqueness argument.
The misconception in applying Theorem 1. The proportion is AD/DB = AE/EC, or equivalently AD/AB = AE/AC. It is not AD/DB = DE/BC. In the model, AD/DB = 2 but DE/BC = 7/10.5 = 2/3. The parallel segment DE compares to BC in the ratio of the whole sides, AD/AB, not in the ratio of the pieces. Write the similarity statement, triangle ADE ~ triangle ABC, and read the ratios from it.
✏️Section 2 · ModelWork through the examples
Worked, Theorem 1 with an unknown: In triangle PQR, S is on PQ and T is on PR with ST parallel to QR. PS = 10, SQ = 4 and PT = 15. Then PS/SQ = PT/TR gives 10/4 = 15/TR, so 10 · TR = 60 and TR = 6. Check with whole sides: PS/PQ = 10/14 = 5/7 and PT/PR = 15/21 = 5/7. If QR = 21, then ST = (5/7)(21) = 15.
Worked, the converse: In triangle ABC, D is on AB with AD = 5 and DB = 7.5, and E is on AC with AE = 4 and EC = 6. AD/DB = 5/7.5 = 2/3 and AE/EC = 4/6 = 2/3. The ratios are equal, so by the converse DE is parallel to BC. If instead EC = 7, then AE/EC = 4/7, which is not 2/3, so DE would not be parallel to BC: the line through D parallel to BC meets AC at the point 4 units from A only when EC = 6.
Worked, Pythagorean pieces: A right triangle has legs 5 and 12, so the hypotenuse is 13. With the altitude to the hypotenuse, the piece next to the leg 5 is 5²/13 = 25/13 ≈ 1.923, the piece next to the leg 12 is 12²/13 = 144/13 ≈ 11.077, and the two pieces add to 169/13 = 13. The altitude is 5 · 12/13 = 60/13 ≈ 4.615, and (60/13)² = 3600/169, which equals (25/13)(144/13) = 3600/169, as the similarity of the two small triangles predicts.
Worked, writing the similarity proof: Right triangle with right angle at C and altitude CD. Triangle ACD ~ triangle ABC because angle A is shared and angles ADC and ACB are both right angles (AA). Hence AC/AB = AD/AC, so AC² = AB · AD. Triangle CBD ~ triangle ABC because angle B is shared and angles CDB and ACB are right angles (AA). Hence BC/AB = BD/BC, so BC² = AB · BD. Adding, AC² + BC² = AB(AD + BD) = AB², since D lies between A and B (angles A and B are acute), so AD + BD = AB.
🧩Section 3 · DoPractice, demonstrate, and explain
Recording task: Draw a right triangle with legs 6 cm and 8 cm and measure its hypotenuse, expecting 10 cm. Draw the altitude from the right angle to the hypotenuse with a set square and measure the two pieces of the hypotenuse and the altitude, expecting 3.6 cm, 6.4 cm and 4.8 cm. Record a table with the three triangles (large, and the two small ones), their angle measures, and the ratios leg/hypotenuse in each, and mark the pairs of similar triangles with the AA reason. Beneath it write the two-line calculation 6² = 10 · 3.6 and 8² = 10 · 6.4 and the sum 36 + 64 = 100.
Independent performance: Write complete proofs of Theorem 1 (a line parallel to one side divides the other two proportionally) and of the Pythagorean Theorem using similarity, each as a numbered list of statements with a reason for every statement, and a proof of the converse of Theorem 1 using the uniqueness of the point that divides a segment in a given ratio. Then apply each: in triangle ABC with D on AB and E on AC and DE parallel to BC, find EC when AD = 9, DB = 6 and AE = 12; decide whether DE is parallel to BC when AD = 6, DB = 4, AE = 9 and EC = 6; and find the two pieces of the hypotenuse and the altitude for a right triangle with legs 9 and 12. Success criteria: every step of each proof cites AA, a parallel-line angle fact, or algebra; the correspondence of vertices in each similarity is written explicitly; the converse proof does not assume what it is proving; and every applied number is checked by a second route (whole-side ratios, or the pieces adding to the hypotenuse).
✅Section 4 · Check7-question lesson practice test
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
Problem 1
In triangle ABC, D is on AB and E is on AC with DE parallel to BC. AD = 8, DB = 4 and AC = 9. Find AE and EC, and state the theorem you used.
Show worked solution
Because DE is parallel to BC, the line divides the other two sides proportionally: AD/AB = AE/AC, so AE = 9 × 8/12 = 6 and EC = 9 − 6 = 3. Check: AD/DB = 8/4 = 2 and AE/EC = 6/3 = 2. The theorem rests on triangle ADE being similar to triangle ABC by AA, using the corresponding angles made by the transversal AB with the parallel lines.
Problem 2
Prove that a line parallel to one side of a triangle divides the other two sides proportionally.
Show worked solution
Let DE be parallel to BC with D on AB and E on AC. Angle ADE = angle ABC (corresponding angles, transversal AB across parallel lines) and angle A is common, so triangle ADE ~ triangle ABC by AA. Hence AD/AB = AE/AC. Taking reciprocals and writing AB = AD + DB, AC = AE + EC gives 1 + DB/AD = 1 + EC/AE, so DB/AD = EC/AE, which is AD/DB = AE/EC.
Problem 3
D is on AB with AD = 6 and DB = 4; E is on AC with AE = 9 and EC = 6. Is DE parallel to BC? Justify with the converse.
Show worked solution
AD/DB = 6/4 = 3/2 and AE/EC = 9/6 = 3/2. The ratios are equal, so by the converse of the side-splitting theorem DE is parallel to BC. The converse is proved by drawing the line through D parallel to BC, noting it divides AC in the ratio 3/2 by the theorem, and observing that only one point of AC divides it in that ratio, so that line passes through E.
Problem 4
A right triangle has legs 9 and 12. Find the two pieces into which the altitude to the hypotenuse divides it, and the altitude, and check the Pythagorean Theorem from the pieces.
Show worked solution
The hypotenuse is √(81 + 144) = 15. The piece next to the leg 9 is 9²/15 = 81/15 = 5.4 and the piece next to the leg 12 is 12²/15 = 144/15 = 9.6; they add to 15. The altitude is 9 × 12/15 = 7.2, and 7.2² = 51.84 = 5.4 × 9.6. Then 9² + 12² = 15 × 5.4 + 15 × 9.6 = 15 × 15 = 225, which is the Pythagorean Theorem obtained from the similar triangles.
Problem 5
Explain why triangle ACD is similar to triangle ABC when CD is the altitude to the hypotenuse of right triangle ABC (right angle at C), and write the proportion that gives AC². Then complete a proof of the Pythagorean Theorem from the two similarities.
Show worked solution
Both triangles contain angle A, and angle ADC and angle ACB are both right angles, so two pairs of angles agree and the triangles are similar by AA, with A ↔ A, C ↔ B, D ↔ C. Corresponding sides give AC/AB = AD/AC, so AC² = AB × AD: the leg squared equals the hypotenuse times the piece of the hypotenuse adjacent to that leg. In the same way triangle CBD ~ triangle ABC by AA (angle B is common, and the angles at D and C are right), with C ↔ A, B ↔ B, D ↔ C, so BC/AB = BD/BC and BC² = AB × BD. Adding, AC² + BC² = AB × (AD + BD). Because angles A and B are acute, the foot D lies between A and B, so AD + BD = AB, and AC² + BC² = AB².
Problem 6
A student writes AD/DB = DE/BC for a segment DE parallel to BC. Using AD = 8, DB = 4, BC = 10.5, show why this is wrong and give the correct value of DE.
Show worked solution
From the similarity ADE ~ ABC, DE/BC = AD/AB = 8/12 = 2/3, so DE = 7. The student’s equation would give DE = 10.5 × 2 = 21, longer than BC itself, which is impossible for a segment inside the triangle. The parallel segment compares to the third side in the ratio of the whole sides, not the pieces.
Problem 7
Why is the converse of the side-splitting theorem a separate statement needing its own proof? Sketch that proof.
Show worked solution
The theorem says parallel implies proportional; the converse says proportional implies parallel, a different claim. Proof of the converse: given AD/DB = AE/EC, draw the line through D parallel to BC and let it meet AC at E*. By the theorem, AD/DB = AE*/E*C, so AE*/E*C = AE/EC. Exactly one point of AC divides it in a given ratio, so E* = E and DE is that parallel line.