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

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

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

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

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

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

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

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

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

$y^2=2x^2$

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

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What are you trying to show? You're assuming Pythagoras Theorem, and then showing that it holds for a specific case? Why not just apply it directly? –  Calvin Lin Dec 30 '12 at 7:53
i just need a proof that shows that the pythagoras theorem holds for an isosceles right triangle, i.e. a symmetrical right triangle. –  Rajesh K Singh Dec 30 '12 at 8:03
Rajesh: If you're assuming the isosceles is a right triangle, then your proof is valid. There may be a shorter proof. –  coffeemath Dec 30 '12 at 8:10

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 Consider a triangle with sides, $3,4,5$, does $3^2+4^2=5^2$ hold for such a triangle. and how do we prove that $a^2+b^2=c^2$ for a right angled triangle., even though those are both several months old.

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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 $2x^2=y^2$.

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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.