Euclid’s Proof of the Pythagorean Theorem

by Adrian Pascual and Graham Calvario

The proof…

So we create a right triangle and three squares as shown in the figure below.

The proof…

Then we construct line DJ which is parallel to CI and GH. Let M be the point where CG and DJ intersect.

(image with line DJ)

The proof…

PART 1

First we need to show that the area of ABCD and CMJI are equal.

The proof…

We construct triangles CBG and CDI. We will show that these two triangles are congruent.

The proof…

Notice that angle BCD and angle MCI

are both 90°

therefore angle BCG and angle DCI are congruent.

The proof…

Moreover, sides BC and CD are congruent because they are the sides of the same square ABCD.

The proof…

Knowing that angles BCG and DCI are congruent, and sides BC and CD are also congruent, therefore sides BG and DI must be congruent as well.

By SAS, ΔCBG and ΔCDI are congruent.

The proof…

By SAS, ΔCBG and ΔCDI are congruent.

So,

½ (BC)(CD) ≡ ½ (CI)(CM)

(CD)(CD)≡(CI)(CM)…(since BC=CD)

(CD) 2 ≡(CI)(CM)

Notice that (CD) 2 = area of ABCD

And (CI)(CM)= area of CMJI

Therefore, the area of ABCD is equal to the area of the CMJI.

The proof…

We have shown that the area of ABCD is equal to the area of the CMJI.

The proof…

PART 2

Now we are going to show that the area of DEFG and GMJH are equal

The proof…

We construct triangles GFC and GDH. We will show that these two triangles are congruent.

The proof…

Notice that angle DGF and angle MGH

are both 90°

Therefore angle BCG and angle DCI

are congruent.

The proof…

Moreover, sides FG and DG are congruent because they are

the sides of the same square

ABCD.

The proof…

Knowing that angles FGC and DGH are congruent, and sides FG and DG are also congruent, therefore sides FC and DH must be congruent as well.

(image of the two congruent triangles)

By SAS, ΔGFC and ΔGDH are congruent.

The proof…

By SAS, ΔGFC and ΔGDH are congruent.

So,

½ (FG)(DG) ≡ ½ (GH)(GM)

(DG)(DG)≡(GH)(GM)…(since FG=DG)

(DG) 2 ≡(GH)(GM)

Notice that (DG) 2 = area of DEFG

And (GH)(GM)= area of GMJH

Therefore, the area of DEFG is equal to the area of the GMJH.

The proof…

FINALLY we merge what we got from Part 1 and Part 2.

Using some algebra we get:

(CD)2 + (DG)2 =(CI)(CM) + (GH) (GM)

(CD)2 + (DG)2 =(CI)(CM) + (CI) (GM) …since GH=CI

(CD)2 + (DG)2 =(CI)(CM + GM)

(CD)2 + (DG)2 =(CI)(CG)…since CM+GM=CG

(CD)2 + (DG)2 =(CG)(CG)…since CI=CG

(CD)2 + (DG)2 =(CG)2

The proof…

(CD)2 + (DG)2 =(CG)2 IS THE PYTHAGOREAN THEOREM

CONGRATULATIONS!

Now you have learned Euclid’s proof of the Pythagorean theorem. Good job!

Click the picture if you want to learn it again!