Use proportional cuts and altitude similarity

Try

In triangle ABC, let D lie strictly between A and B and E strictly between A and C. Suppose DE is parallel to BC. If AB=10, AC=15 and AD=4, predict AE. The small triangle shares angle A, and parallel lines match the other angles, so the same scale applies on both sides: AE/15=4/10 and AE=6.

State which lengths are whole sides and which are parts. DB=6 and EC=9. The fraction AD/AB is 4/10, whereas the part-to-part ratio AD/DB is 4/6. They express related information but are not interchangeable numbers.

Predict

Predict the converse. If AD/AB=AE/AC for interior points D and E, must DE be parallel to BC? It is tempting to say yes merely because the first theorem feels reversible. We will instead construct a known parallel through D and prove that its intersection with AC is exactly E.

Also predict why “both sides are shortened by the same amount” is insufficient. If AB=10 and AC=15, removing 2 from each leaves ratios 8/10 and 13/15, which differ. Similarity preserves a common multiplier, not an additive decrement.

Experiment

Draw the 10-and-15 configuration with a chosen nonzero angle at A. Mark D at 4/10 of AB and E at 6/15 of AC. Join D to E and compare its direction with BC. Then move E to a different fraction while leaving D fixed; the parallelism should fail.

Measurement	Along AB	Along AC
Whole side	10	15
Part from A	4	6
Remaining part	6	9
Part divided by whole	2/5	2/5
Part divided by remainder	2/3	2/3

The drawing is a construction to inspect, not the proof. Your written argument must work for arbitrary positive side lengths and interior cut points.

Understand

For the forward theorem, parallel DE and BC give equal corresponding angles, so triangles ADE and ABC are similar by the previously derived AA criterion. Their correspondence is A↔A, D↔B, E↔C. Therefore AD/AB=AE/AC=DE/BC. If the common part-to-whole ratio is r, with 0 < r < 1, then DB=(1−r)AB and EC=(1−r)AC. Dividing gives AD/DB=AE/EC=r/(1−r).

For the converse, assume AD/AB=AE/AC. Draw the unique Euclidean parallel to BC through D. It meets segment AC at an interior point E₀. By the forward theorem, AE₀/AC=AD/AB. The given equality then implies AE₀=AE. Both E and E₀ lie on the same ray from A, so uniqueness of a point at a specified distance gives E₀=E. Hence DE is the constructed parallel. If the hypothesis is instead AD/DB=AE/EC, cross-multiplying and adding the matching product converts it to AD/(AD+DB)=AE/(AE+EC), the required part-to-whole equality.

Now let ABC be a right triangle with right angle at C. Drop the perpendicular from C to hypotenuse AB, meeting it at D. Put AC=u, BC=v, AB=c, AD=p and DB=q. The foot lies inside the hypotenuse, and p+q=c. Triangle ACD is similar to ABC: angle A is common and the right angle at D matches the right angle at C. Matching hypotenuse to hypotenuse and the side beside A gives u/c=p/u, hence u²=cp. The other small triangle gives v/c=q/v, hence v²=cq. Adding yields u²+v²=c(p+q)=c². This is the similarity proof of Pythagoras, distinct from a square-dissection proof.

Apply

Use the right triangle with legs 6 and 8 and hypotenuse 10 as a numerical check of the derived relations. They give p=36/10=18/5 and q=64/10=32/5, which add to 10. The following equal-scale diagram shows those projections. Its coordinates illustrate the established theorem; they are not used as a premise in the general proof.

For a fresh proof, write the vertex correspondence of each small triangle before forming a proportion. A wrong pairing such as u/c=q/u generally fails because q lies next to the other endpoint. Every ratio must connect corresponding sides, not merely convenient letters.

Continue

The parallel-cut theorem and its converse connect construction with measurement: a parallel produces proportional parts, and proportional parts can certify a parallel. The Pythagorean proof uses those similarity ideas to connect the two legs with the hypotenuse.

Optional depth: the two small triangles are also similar to each other, giving altitude squared equal to pq. Derive that correspondence separately. Do not replace the argument with the numerical fact that it happens to hold in the 6–8–10 example.