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Question Number 218236 by mr W last updated on 02/Apr/25

Commented by mr W last updated on 02/Apr/25

$$whyisthisquestiondeleted?$$

Commented by mr W last updated on 02/Apr/25

$$igotx=15°$$

Commented by mr W last updated on 02/Apr/25

Answered by mr W last updated on 02/Apr/25

Commented by mr W last updated on 02/Apr/25

$$let\mathrm{\Delta }BEA\equiv \mathrm{\Delta }BDC$$$$\mathrm{\angle }EBD=90°$$$$\mathrm{\angle }ADE=180°-2x-(90°-3x)-45°=45°+x$$$$\mathrm{\angle }DAE=2x+45°-x=45°+x=\mathrm{\angle }ADE$$$$\Rightarrow DE=AE=DC$$$$\mathrm{\angle }CDE=45°+180°-3x-(45°-x)=180°-2x$$$$\Rightarrow \mathrm{\angle }DCE=\mathrm{\angle }DEC=\frac{180°-(180°-2x)}{2}=x$$$$\mathrm{\angle }CEA=180°-3x-(45°-x)-45°+x=90°-x$$$$\mathrm{\angle }CAE=45°-x+45°=90°-x=\mathrm{\angle }CEA$$$$\Rightarrow CA=CE\Rightarrow AD=ED$$$$\Rightarrow \mathrm{\Delta }ADEisequilateral$$$$\Rightarrow 45°+x=60°$$$$\Rightarrow x=15°$$

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