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pythagoras reciprocal theorem lema

I'm trying to understand the proof of the reciprocal Pythagorean theorem, but it seems to depend on the fact that $ab = cd$ in the picture above, but I cannot see why. Assuming the result, I can then apply Pythagoras getting $a^2 + b^2 = c^2$ which implies $1/b^2 + 1/a^2 = c^2/(a^2b^2)$ which implies $1/b^2 + 1/a^2 = c^2/(c^2d^2)$ which implies $1/b^2 + 1/a^2 = c^2/(c^2d^2)$ ending up with $1/b^2 + 1/a^2 = 1/d^2$, as desired. (So I'm missing only why $ab = cd$.)

JMP
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user1145880
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    Consider the area in two different ways, with two different bases. – Vivaan Daga Feb 02 '23 at 12:13
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    This is necessary for the condition that the triangle has a right angle. In this case , both $ab$ and $cd$ are half of the area of the triangle. – Peter Feb 02 '23 at 12:13
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    It's the area of a rectangle spanned by c and d. This area is equal to the span of the parallelogram spanned by a and b – justabit Feb 02 '23 at 12:14
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    @Peter You mean they are twice the triangle area. – coffeemath Feb 02 '23 at 12:20
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    @coffeemath Oh, of course ! Humiliating ... – Peter Feb 02 '23 at 12:21
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    Oh, I finally see it. I can consider the big triangle's area in two different ways: one where $b$ is the height and $a$ is the base --- so I get its area equal to $ab/2$ --- and the other where $d$ is the height and $c$ is the base --- so I get $cd/2$. Equating both, I get $ab = cd$. Phew. Simple, huh? I had not seen it, despite the hint in the image. Thanks very much. – user1145880 Feb 02 '23 at 12:26

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Let the angle between $a$ and $c$ be $\alpha$. Note that $\sin(\alpha)=\cos(90^\circ-\alpha)$.

Therefore $\sin(\alpha)=\frac{d}{a}$, so $a=\frac{d}{sin(\alpha)}$.

Similarly, $\cos(90^\circ-\alpha)=\sin(\alpha)=\frac{b}{c}$, so $b=c\sin(\alpha)$.

Therefore $ab=cd$.

JMP
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