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Question Number 199023 by mr W last updated on 26/Oct/23

Commented by mr W last updated on 26/Oct/23

$$ifAX=YB,prove\mathrm{\angle }ADX=\mathrm{\angle }YDC.$$

Answered by mr W last updated on 27/Oct/23

Commented by mr W last updated on 27/Oct/23

$$AD=YO=radius$$$$\mathrm{\Delta }ADX\equiv \mathrm{\Delta }YOB$$$$\mathrm{\angle }ADX=\mathrm{\angle }YOB=\alpha =\mathrm{\angle }COB$$$$\mathrm{\angle }YDC=\beta =\frac{\mathrm{\angle }YOC}{2}=\alpha =\mathrm{\angle }YOB=\mathrm{\angle }ADX$$$$\Rightarrow proved$$

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