What follows is a sequence of guided student exercises.
(a) Given similar right triangles ABC and DEF below, let b_{1} be the length of AB, b_{2} that of DE, h_{1} that of CB, and h_{2} that of FE (Figure 6).
Figure 6: Similar Right Triangles.
By definition of similar triangles, ∠A \(\cong\) ∠D, ∠B \(\cong\) ∠E, ∠C \(\cong\) ∠F. For part (a), consider as given that both ∠B and ∠E are right angles. Prove that

by the following outline. Place point D on point C so that the points A, C, and F are collinear (Figure 7).
Figure 7: Arrangement of the Triangles.
Duplicate triangle ABC to form triangle AJC, and duplicate triangle DEF to form triangle DLF, pictured below (Figure 8).
Figure 8: The Duplicated Triangles.
Why are ∠JAB and ∠EFL both right angles? Explain why lines AJ and FL, if extended, will meet in a right angle. Also, explain why lines AB and FE, if extended, will meet in a right angle. Let K and M be these two points of intersection, respectively. Sketch the resulting rectangle AKFM, and apply the inclusionexclusion principle to find an algebraic expression for the area of the subrectangles KLCJ and DEMB in terms of b_{1}, b_{2}, h_{1}, h_{2}. What can be concluded from the resulting equation?
(b) In triangle ABC and triangle DEF (Figure 6) let d_{1} be the length of AC and d_{2} the length of DF. Use the gougu (Pythagorean) theorem and the result of part (a) to prove algebraically that

(c) Develop a version of the inclusionexclusion principle that applies to an arbitrary parallelogram ABCD (Figure 9).
Figure 9: Parallelogram ABCD .
Begin by constructing diagonal DB and choosing a point H along the diagonal.
(d) Let triangle ABC and triangle DEF be arbitrary similar triangles (not necessarily right triangles) with ∠A \(\cong\) ∠D, ∠B \(\cong\) ∠E, ∠C \(\cong\) ∠F (Figure 10).
Figure 10: Arbitrary Similar Triangles.
Let b_{1} be the length of AB, b_{2} the length of DE, s_{1} the length of CB, and s_{2} the length of FE. Prove that

using either of the strategies below.
(e) Prove the gougu (Pythagorean) theorem as another application of the inclusionexclusion principle, following the footsteps of the ancient Chinese. Given right triangle ABC with ∠B a right angle, let a be the length of AB, b the length of CB, and c the length of AC. We wish to prove that

First consider the case of an isosceles right triangle, where a = b. By duplicating triangle ABC four times, construct square ACDE (Figure 11).
Figure 11: Square ACDE with a = b.
Explain why the area of this square is c^{2}. Using the area of triangle ABC, explain why square ACDE also has area

Now consider right triangle ABC with b > a. Again construct square ACDE with side length c (Figure 12).
Figure 12: Square ACDE with b > a.
Explain why the inside square has side length b  a. By excluding this inside square from the outside square, find an expression for

using two methods, one algebraic, the other geometric in terms of the area of the four remaining triangles. What results after equating these two expressions?
Extra Credit A: Verify the chong cha method by demonstrating that in Figure 3, we have

where h is the length of AS, d the length of SN, a_{1} the length of SB, and a_{2} the length of ND. Hint: Draw line CK parallel to AB, and consider two pairs of similar triangles, triangles PRA and ASB and triangles PAC and CKD.
Extra Credit B: Read and study Proposition I.43 from the first book of Euclid's Elements. Compare Euclid's use of rigor in the proof of I.43 to the more intuitive application of the "inout" principle for the rectangle in Figure 5. Speculate why Euclid did not use I.43 to prove VI.4. What may have prevented the ancient Greeks from developing an algebra of real numbers that would have allowed the transfer of results like I.43 to situations like VI.4? What obstacles do incommensurable magnitudes pose in the construction of a real number system?