How do I show that the pythagoras theorem holds for the specific case of an "isosceles right triangle"?

Question

Figure shows a rectangle $ABCD$ and an isosceles triangle $\triangle DEC$.

$AD=BC=z$;$AB=DC=y$;$DE=CE=x$

One solution is as follows.

We know that the pythagoras theorem holds for a right triangle $\triangle EBC$.

In the right triangle, $\triangle EBC$.

$z^2+(\frac{y}{2})^{2}=x^2$ ----(|)

We also know that, AREA of $\triangle DBC$ = AREA of $\triangle DEC$.

$\frac{1}{2}(y)(z)=\frac{1}{2}(x)(x)$ ----(||)

We need to show, $y^2=x^2+x^2$

From, (|) and (||) we have, $z=\frac{x^2}{y}$, $x^2=\frac{y^2}{4}+z^2$

Now, $x^2=\frac{y^2}{4}+\frac{x^4}{y^2}$

$(4x^2)y^2=y^4+4x^4$

$y^4-(4x^2)y^2+4x^4=0$

$y^4-((2x^2)y^2+(2x^2)y^2)+4x^4=0$

$y^4-(2x^2)y^2-(2x^2)y^2+4x^4=0$

$y^2(y^2-2x^2)-(2x^2)(y^2-2x^2)=0$

$(y^2-2x^2)^2=0$

$y^2-2x^2=0$

$y^2=2x^2$

$y^2=x^2+x^2$ ----(|||)

Answer 1

Step 1: Assume the Pythagorean theorem for right triangles.

Step 2: Realize that an isosceles right triangle is a right triangle.

Step 3: Conclude that the Pythagorean theorem for isosceles right triangles is a special case (and therefore free from) the Pythagorean theorem for right triangles.

In addition, I can't help but notice the similarity between this and your previous questions https://math.stackexchange.com/q/182332/9754 and https://math.stackexchange.com/q/184003/9754, even though those are both several months old.

Answer 2

In your drawing add the midpoint of the line AB and call it F. The areas of all the following triangles are equal: AED, BCE,DEF,CFE. So the area of the triangle DEC is halt the area of the rectanle ABCD. But the triangle DEC is half of the square with the side of length x and the rectangle ABCD is half of the square with side y, therefore $2x^2=y^2$.

Answer 3

If I understood you correctly, what you want is a proof of pythagorean theorem for the isosceles right triangle. A proof is as follows, and is similar to a proof of the generalized pythagorean theorem.

Let $ABCD$ be a square of side length $2l$. Let $E, F, G, H$ be the midpoints of $AB, BC, CD, DA$ respectively. By rotational symmetry, $EFGH$ is a square of unknown length $s$. Consider the area of $ABCD$. It is clearly equal to $2l * 2l = 4l^2$. By using the decomposition, $ABCD$ is the sum of the 4 triangles and the square, so has areas $4 \times \frac {1}{2} \times l \times l$, and the square has area $s^2$. Hence,

$$ 4l^2 = 4 \times \frac {1}{2} l^2 + s^2 \Rightarrow l^2 + l^2 = s^2$$

Of course, the preferable way is to prove Pythagorean theorem directly, using a square of side length $a+b$, and letting E, F, G, H, cut the sides into $a$ and $b$. Try doing this on your own.

Answer 4

So, let c be the hypotenuse, and a the common leg-length.

First, note that given a length for the hypotenuse, any two isosceles right triangles having that hypotenuse will be congruent.

Then, use that fact to put the special case of Bhaskara’s proof to work, by drawing a square with side-length c, and then drawing the two diagonals. This exactly divides the square up into 4 copies of the isosceles right triangle in question, each of which has area (a^2)/2. There being 4 of them means that the area of the square, is 2(a^2). So, since c^2 is also the area of the square, we have that c^2 = a^2 + a^2. Done