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Question Number 218527 by Spillover last updated on 11/Apr/25

Answered by A5T last updated on 12/Apr/25

$$\mathrm{CD}=\mathrm{BD}\wedge \mathrm{\angle }\mathrm{CDB}=90°\Rightarrow \mathrm{BC}=\mathrm{CD}\sqrt{2}=2\mathrm{R}$$$$\Rightarrow \mathrm{CD}=\mathrm{R}\sqrt{2}$$$${\mathrm{Ptolemy}}^{\prime }\mathrm{s}\mathrm{theorem}:\mathrm{CD}\times \mathrm{AB}+\mathrm{AC}\times \mathrm{BD}=\mathrm{AD}\times \mathrm{BC}$$$$\Rightarrow (\mathrm{AB}+\mathrm{AC})=\mathrm{AD}\sqrt{2}...\left(\mathrm{i}\right)$$$${\mathrm{AB}}^2+{\mathrm{AC}}^2={\mathrm{BC}}^2$$$$\Rightarrow {(\mathrm{AB}+\mathrm{AC})}^2-2\times \mathrm{AB}\times \mathrm{BC}={4\mathrm{R}}^2$$$$\left(\mathrm{i}\right)\Rightarrow {2\mathrm{AD}}^2-4\left[\mathrm{ABC}\right]={4\mathrm{R}}^2$$$$\Rightarrow \left[\mathrm{ABC}\right]=\frac{{\mathrm{AD}}^2}{2}-{\mathrm{R}}^2$$

Commented by Spillover last updated on 17/Apr/25

$$thankyou.$$

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