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Question Number 211190 by mr W last updated on 30/Aug/24

Answered by A5T last updated on 31/Aug/24

$$Let\mathrm{\angle }BCD=\mathrm{\angle }CBD=\theta \Rightarrow \mathrm{\angle }DBA=90°-\theta$$$$\Rightarrow \mathrm{\angle }ADB=90°-\theta \Rightarrow \mathrm{\angle }DAB=2\theta$$$$\mathrm{\angle }BCA=\mathrm{\angle }CAB=45°\Rightarrow \mathrm{\angle }ACD=45°-\theta \Rightarrow ?=45°-2\theta$$$$LetCD=BD=xandAD=AB=BC=y$$$$Then{AC}^2=2y^2=x^2+y^2-2xycos(3\theta +90°)...\left(i\right)$$$${BD}^2=x^2=x^2+y^2-2xycos\left(\theta \right)\Rightarrow y=2xcos\theta ...\left(ii\right)$$$$\left(ii\right)in\left(i\right)\Rightarrow 4{cos}^2\theta =1-4\left(cos\theta \right)cos(3\theta +90)$$$$\Rightarrow \theta =15°\Rightarrow ?=45°-2\theta =15°$$

Commented by mr W last updated on 31/Aug/24

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Answered by mm1342 last updated on 31/Aug/24

$$<DBC=\theta \Rightarrow <ADB=90-\theta \mathrm{\&}<CDB=180-2\theta \mathrm{\&}<DAB=2\theta$$$$\frac{CD}{BC}=\frac{sin\theta }{sin2\theta }=\frac{1}{2cos\theta }\mathrm{\&}\frac{BD}{AD}=\frac{sin2\theta }{cos\theta }=2sin\theta$$$$\Rightarrow sin2\theta =\frac{1}{2}\Rightarrow \theta =15✓$$$$$$

Commented by mr W last updated on 31/Aug/24

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Answered by mr W last updated on 31/Aug/24

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