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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

proof for Q138519

$${proof}\:{for}\:{Q}\mathrm{138519} \\ $$

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

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

α+β+γ=360°  ⇒(α/2)+(β/2)+(γ/2)=180°  that means triangles with interior  angles (α/2),(β/2),(γ/2) can be formed, such  as ΔP′Q′R′, see diagram above.  the circles with centers at P,Q,R   intersect at point D.  ∠DPQ=((∠DPC)/2)  ∠DPR=((∠DPB)/2)  ∠P=∠DPQ+∠DPR=((∠DPC+∠DPB)/2)=(α/2)  similarly  ∠Q=(β/2)  ∠R=(γ/2)

$$\alpha+\beta+\gamma=\mathrm{360}° \\ $$$$\Rightarrow\frac{\alpha}{\mathrm{2}}+\frac{\beta}{\mathrm{2}}+\frac{\gamma}{\mathrm{2}}=\mathrm{180}° \\ $$$${that}\:{means}\:{triangles}\:{with}\:{interior} \\ $$$${angles}\:\frac{\alpha}{\mathrm{2}},\frac{\beta}{\mathrm{2}},\frac{\gamma}{\mathrm{2}}\:{can}\:{be}\:{formed},\:{such} \\ $$$${as}\:\Delta{P}'{Q}'{R}',\:{see}\:{diagram}\:{above}. \\ $$$${the}\:{circles}\:{with}\:{centers}\:{at}\:{P},{Q},{R}\: \\ $$$${intersect}\:{at}\:{point}\:{D}. \\ $$$$\angle{DPQ}=\frac{\angle{DPC}}{\mathrm{2}} \\ $$$$\angle{DPR}=\frac{\angle{DPB}}{\mathrm{2}} \\ $$$$\angle{P}=\angle{DPQ}+\angle{DPR}=\frac{\angle{DPC}+\angle{DPB}}{\mathrm{2}}=\frac{\alpha}{\mathrm{2}} \\ $$$${similarly} \\ $$$$\angle{Q}=\frac{\beta}{\mathrm{2}} \\ $$$$\angle{R}=\frac{\gamma}{\mathrm{2}} \\ $$

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