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Question Number 155056 by mathdanisur last updated on 24/Sep/21

Answered by mr W last updated on 24/Sep/21

$${BF}^2={AB}^2+{\left(\sqrt{2}AC\right)}^2+2\times AB\times \sqrt{2}AC\times \cos (A+\frac{\pi }{4})..\left(i\right)$$$${CD}^2={AC}^2+{\left(\sqrt{2}AB\right)}^2+2\times AC\times \sqrt{2}AB\times \cos (A+\frac{\pi }{4})..\left(ii\right)$$$$\left(i\right)-\left(ii\right):$$$${BF}^2-{CD}^2={AC}^2-{AB}^2$$$$\Rightarrow {AB}^2+{BF}^2={AC}^2+{CD}^2$$

Commented by mathdanisur last updated on 24/Sep/21

$$\mathrm{very}\mathrm{nice}\mathrm{Ser}\mathrm{thank}\mathrm{you}$$

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