Name(s):
Unlike dissection proofs, in which you move pieces to different locations in your figure without changing them, in this construction you’ll transform your squares, without changing their areas, to create congruent shapes. To do this, you’ll actually construct parallelograms on the sides of your right triangle.
Sketch and Investigate
1. Construct a right triangle and squares on the sides.
2. Hide the far side of the square on side a and replace it with a line.
B
In steps 3 through 5, you’ll construct a parallelogram on side a .
This transformation is called a shear.
A shear translates every point in a figure in a direction parallel to a given line by a distance proportional to the point’s distance from the line. Shearing a figure preserves its area.
3. Construct CF , where F is a point on the line.
4. Construct a line through B , parallel to CF .
G
F
C a b c
A
5. Construct the interior of parallelogram BCFG and hide the lines.
Drag point F to experiment with the parallelogram’s behavior. The key to this demonstration is that the parallelogram’s area is always the same as the square whose side it shares. Can you see why?
6. Construct a parallelogram on side b by the same method you used in steps 2 through 5 above, starting with segment CH so that H is a point that can be dragged.
G
B
C a b c
A
F
38 • Chapter 2: Proofs and Demonstrations
H J
Pythagoras Plugged In: Proofs and Problems for The Geometer’s Sketchpad
© 2003 Key Curriculum Press

A Dynamic Proof (continued)
You should now have parallelograms on sides a and b that can be dragged back and forth by one vertex without changing their areas.
7. Construct a line through point C , perpendicular to side c .
L K'
Note that this concave hexagon consists of two parallelograms,
BLK´K and AKK´M.
In your sketch , you may need to drag a different point , depending on how you constructed the parallelogram.
8. Construct BK and AK , where K is a point on this line.
B K
M
9. Mark BL as a vector and translate BK , point K , and
AK by this marked vector.
G a c
C b
10. Construct the polygon interior BLK´MAK.
A
F
11. Drag point K to make sure polygon BLK´MAK behaves correctly: Point K´ should move when you drag K so that BLK ´ K and AKK´M remain parallelograms.
H J
If everything is working properly, you’re ready to manipulate your sketch to give a dynamic proof of the Pythagorean theorem.
12. Drag point F to shear the parallelogram back and forth. Note that the parallelogram can fill the square. Note too that the parallelogram’s area doesn’t change. (You can measure to confirm this.) Finish the shear by dragging F to lie on the line through C and K .
Q1 Explain why the area of the parallelogram doesn’t change.
13. Shear the parallelogram on the other leg of the triangle. Note that it too fills its square and that its area doesn’t change. Drag until point H is on the line.
14. Drag point K to shear the parallelograms in the square on c . Note that these parallelograms can fill the square and that the area of the polygon (the sum of the areas of the parallelograms) doesn’t change.
Drag until point K coincides with point C .
Q2 The figures on the next page show the sequence you might have performed. How does this demonstrate the Pythagorean theorem?
Pythagoras Plugged In: Proofs and Problems for The Geometer’s Sketchpad
© 2003 Key Curriculum Press
Chapter 2: Proofs and Demonstrations • 39

A Dynamic Proof (continued)
Stage 1 Stage 2
Stage 3 Stage 4
Prove
The dynamic “proof” you perform by manipulating your sketch to transform the squares into two congruent figures may seem proof enough of the Pythagorean theorem. People might consider the details of the proof superfluous compared to the convincing power of the dynamic demonstration. Yet if you were showing this proof to someone, you’d probably want to explain what’s going on and why your sketch proves the theorem. And to make it a complete, or rigorous, proof, you’d want to supply the details. Start with a figure like that in stage 4 above. Add labels and write a complete proof. That is, show that each parallelogram is equal in area to a square (or part of a square) and show that the two pairs of parallelograms are congruent.
40 • Chapter 2: Proofs and Demonstrations Pythagoras Plugged In: Proofs and Problems for The Geometer’s Sketchpad
© 2003 Key Curriculum Press

A Dynamic Proof (continued)
Explore More
Write a paragraph about any similarities you see between this proof and
Euclid’s proof.
A Presentation Sketch
This sketch lends itself nicely to being made into a presentation sketch — a sketch whose actions can be driven by double-clicking action buttons. In the figure below, point C ´ is the translated image of point C by vector LB .
Make the action buttons indicated. The Shear button is a Presentation button that sequences the first three Move buttons. The Return to Squares button presents the second set of three Move buttons simultaneously.
Once the Presentation buttons are made, hide the separate Move buttons.
L
G
J
F
B
C
N
K
C' H K
Shear Return to Squares
Move F -> C'
Move H -> C'
Move K -> C
Move F -> J
Move H -> K
Move K -> N
Pythagoras Plugged In: Proofs and Problems for The Geometer’s Sketchpad
© 2003 Key Curriculum Press
Chapter 2: Proofs and Demonstrations • 41

A Dynamic Proof
Sketchpad Proficiency : Advanced
Example Sketch : Dynamic.gsp
Sketch and Investigate
Consider not giving students the third page of the activity at first to see if they can discover for themselves how to demonstrate the Pythagorean theorem with this figure.
Q1 The area of the parallelogram doesn’t change because neither its base nor its height changes.
Prove
L
I
B
C a
D c b
K'
A
G
M
F N
J
Parallelogram BCFG has base a and height a , so its area is a 2 . Parallelogram ACFJ has base and height b , so its area is b
2
. Parallelograms BCK´L and
ACK´M have bases and heights (and thus areas) equal to the bases and heights of rectangles BDIL and ADIM , respectively. So the sum of the parallelograms’ areas is the sum of the areas of the rectangles. The rectangles comprise the square on c , so the sum of the areas of BCK´L and ACK´M is c
2
.
∆
AJN is congruent to
∆
ABC , so AJ = c = AM . The four angles of parallelogram ACFJ are congruent to the angles of parallelogram ACK´M , and they share side AC , so the parallelograms are congruent. Likewise FC and CK´ in parallelograms
GFCB and BCK´L equal c , the parallelograms share side BC , and their corresponding angles are congruent, so GFCB and BCK´L ´ are congruent.
Hence, area GFCB + area ACFJ = area BCK´L + area ACK´M, and a
2
+ b
2
= c
2
.
Pythagoras Plugged In: Proofs and Problems for The Geometer’s Sketchpad
© 2003 Key Curriculum Press
Activity Notes • 77