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Question Number 138661 by mr W last updated on 16/Apr/21

Commented by mr W last updated on 16/Apr/21

$$proofforQ138519$$

Commented by mr W last updated on 16/Apr/21

Commented by mr W last updated on 16/Apr/21

$$\alpha +\beta +\gamma =360°$$$$\Rightarrow \frac{\alpha }{2}+\frac{\beta }{2}+\frac{\gamma }{2}=180°$$$$thatmeanstriangleswithinterior$$$$angles\frac{\alpha }{2},\frac{\beta }{2},\frac{\gamma }{2}canbeformed,such$$$$as\mathrm{\Delta }{P}^{\prime }{Q}^{\prime }{R}^{\prime },seediagramabove.$$$$thecircleswithcentersatP,Q,R$$$$intersectatpointD.$$$$\mathrm{\angle }DPQ=\frac{\mathrm{\angle }DPC}{2}$$$$\mathrm{\angle }DPR=\frac{\mathrm{\angle }DPB}{2}$$$$\mathrm{\angle }P=\mathrm{\angle }DPQ+\mathrm{\angle }DPR=\frac{\mathrm{\angle }DPC+\mathrm{\angle }DPB}{2}=\frac{\alpha }{2}$$$$similarly$$$$\mathrm{\angle }Q=\frac{\beta }{2}$$$$\mathrm{\angle }R=\frac{\gamma }{2}$$

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