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Question Number 61952 by lalitchand last updated on 12/Jun/19

Commented by lalitchand last updated on 12/Jun/19

$$\mathrm{question}\mathrm{number}3\mathrm{prove}\mathrm{that}$$

Answered by MJS last updated on 12/Jun/19

$$\mathrm{\angle }PAC=180°-\alpha \Rightarrow \mathrm{\angle }CAQ=90°-\frac{\alpha }{2}$$$$\mathrm{\angle }RCA=180°-\gamma \Rightarrow \mathrm{\angle }ACQ=90°-\frac{\gamma }{2}$$$$\Rightarrow \mathrm{\angle }AQC=\frac{\alpha +\gamma }{2}$$$$\beta =180°-(\alpha +\gamma )\Rightarrow (\alpha +\gamma )=180°-\beta$$$$\Rightarrow \mathrm{\angle }AQC=90°-\frac{\beta }{2}$$

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