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Question Number 181437 by mnjuly1970 last updated on 25/Nov/22

Answered by MJS_new last updated on 25/Nov/22

$$\mathrm{let}d,\sqrt{b^2+d^2}\mathrm{the}\mathrm{other}\mathrm{sides}\mathrm{of}\mathrm{the}\mathrm{rectangular}$$$$\mathrm{triangle}$$$$p+q=\sqrt{b^2+d^2};h^2+p^2=d^2\wedge h^2+q^2=b^2$$$$\Rightarrow p=\frac{d^2}{\sqrt{b^2+d^2}}\wedge q=\frac{b^2}{\sqrt{b^2+d^2}}$$$$\Rightarrow a^2={\left(\frac{p}{2}\right)}^2+y^2\wedge c^2={(\frac{p}{2}+q)}^2+y^2$$$$\iff a^2=\frac{d^4}{4(b^2+d^2)}+y^2\wedge c^2=\frac{{(2b^2+d^2)}^2}{4(b^2+d^2)}$$$$\Rightarrow$$$$a^2+b^2=\frac{d^4}{4(b^2+d^2)}+y^2+b^2=$$$$=\frac{d^4}{4(b^2+d^2)}+y^2+\frac{4b^2(b^2+d^2)}{4(b^2+d^2)}=$$$$=\frac{4b^4+4b^2{d}^2+d^4}{4(b^2+d^2)}+y^2=\frac{{(2b^2+d^2)}^2}{4(b^2+d^2)}+y^2=c^2$$

Answered by HeferH last updated on 25/Nov/22

Commented by HeferH last updated on 25/Nov/22

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