Proof that rectangle triangle is isosceles

Question

I would like to proof that XYZ rectangle triangle with legs x, e and e and hypothenuse z with area $\frac{z^2}{4}$ is isosceles. What I tried is to equal $\frac{z^2}{4}$ with $\frac{b.h}{2}$ getting $b$ and $h$ from Pythagoras theorem $a^2+b^2=c^2$ . I was told this is right but doesn't work with rectangle triangles. Any suggestions?

Answer 1

HINT: Let $b=z\cos t,h=z\sin t$

$$\implies \dfrac{z^2}4=\dfrac{z\cos t\cdot z\sin t}2\implies\sin2t=1\implies t=?$$

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Alternatively,

$$z^2=b^2+h^2\ge2bh\implies\dfrac{z^2}4\ge\dfrac{bh}2$$

We need the equality which occurs if $b=h$