Geometry – Finding Ratio of Sides of 45-45-90 Triangle Without Pythagoras Theorem

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I've managed to find out the ratio of sides of $30$–$60$–$90$ triangle through the length of angle bisector, which does not depend on Pythagorean theorem at all, looking at its proof.

I did it in the following way: glue together two equal $30$–$60$–$90$ triangles so their longer legs are joined together to get an isosceles (actually equilateral) triangle. Then that leg is also the angle's bisector by the corresponding property of an isosceles triangle (it is an altitude to the base of isosceles triangle, so it also is a median and a bisector). Let the shorter leg be of length $1$, then the hypotenuse is of length $2$ (an isosceles triangle with angle $60^\circ$ is equilateral, and the shorter leg is half its side), and the remaining leg can be found by the length of angle bisector formula: $\sqrt{2\cdot 2 – 1\cdot 1} = \sqrt{3}$. Thus we have found the ratio of the sides, which is $1:\sqrt{3}:2$.

How do I approach another famous triangle, $45$–$45$–$90$? I'm aware I should use something similar, but no idea how.

These are two drawings I came up with that may possibly lead to some clue.

Using known 30 60 90 ratios
An altitude on hypotenuse splits 45 45 90 into two 45 45 90s

Best Answer

a square with its diagonals and center point: one edge of the square is marked "1", and one half of a diagonal is marked "b"; also, a rectangle, cut into two squares, each with one of its diagonals: one of the diagonals is marked 1, and one of the cut parts of the long edge is marked "b" These two figures are made of four $45-45-90$ triangles, with the same sides and angles. The first produces a square, so its area is $1\times1=1$; the second produces a rectangle, so its area is $b\times2b=2b^2$, but since it's still four triangles just like the other, it has the same area as the first figure, which is to say $1$. so $2b^2=1$, or $b^2 = 1/2$, or $b = \sqrt{1/2}$.

Alternatively, using LoAB:

the double-ticked sides are length $1$, so $d = 1$, $a = 2$, and $b = c$. This gives:

$$\begin{align} d^2&=\frac{bc}{(b+c)^2}\left((b+c)^2-a^2\right)\\ 1^2&=\frac{bb}{(b+b)^2}\left((b+b)^2-2^2\right)\\ 1&=\frac{b^2}{(2b)^2}\left((2b)^2-4\right)\\ 1&=\frac{4b^2-4}{4}\\ 1&=b^2-1\\ 2&=b^2\\ b&=\sqrt{2}\\ \end{align}$$

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