We can prove the in two different ways – by using areas of triangles, squares, and rectangles, or by using similarity properties.
Using Areas
We begin by considering a .
Next, we will consider three more right triangles with the same dimensions and we will arrange them in such a way that we get a large square formed by the legs of the triangles.
From the above, we can see that the area of the large square is equal to four times the area of the initial triangle plus the area the inner square.
A=4⋅A△+A□
The is
c2 and the is
21ab.
A=4⋅21ab+c2⇕A=2ab+c2
On the other side, we can use the lengths of the legs of the triangles to divide the large square into two squares and two rectangles as shown below.
With this division, we can write a second expression for the area of the large square.
A=a2+ab+b2+ab⇕A=a2+b2+2ab
Finally, let's equate the two expressions we got for the area of the large square.
A=2ab+c2
a2+b2+2ab=2ab+c2
a2+b2=c2
As we can see, we have obtained the relation established in the Pythagorean Theorem.
Using Similar Triangles
As before, we begin by considering a right triangle.
Next, we will consider the to the hypotenuse.
From the above, we obtain two new triangles. Let's separate them and mark the congruent angles between them.
By applying the , we get that △ABC∼△ACD and △ABC∼△CBD. Then, we can write the following relations.
△ABC∼△ACD
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△ABC∼△CBD
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bc=c−db=ha
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ac=hb=da
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From the left-hand side column, we will pick the first and second fractions and from the right-hand side column we will pick the first and third fractions.
bc=c−db⇓c(c−d)=b2andac=da⇓cd=a2
Next, let's simplify the left-hand side equation.
c(c−d)=b2
c2−cd=b2
a2+b2=c2
One more time, we've got the relation established in the Pythagorean Theorem.