Let-A-B-and-C-be-points-on-the-sides-BC-CA-and-AB-of-the-triangle-ABC-Prove-that-the-circumcircles-of-the-triangles-AB-C-BA-C-and-CA-B-have-a-common-point-Prove-that-the-property-holds-even

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Question Number 16071 by Tinkutara last updated on 21/Jun/17

$$\mathrm{Let}{A}^{\prime },B^{\prime }\mathrm{and}{C}^{\prime }\mathrm{be}\mathrm{points}\mathrm{on}\mathrm{the}\mathrm{sides}$$$$BC,CA\mathrm{and}AB\mathrm{of}\mathrm{the}\mathrm{triangle}ABC.$$$$\mathrm{Prove}\mathrm{that}\mathrm{the}\mathrm{circumcircles}\mathrm{of}\mathrm{the}$$$$\mathrm{triangles}{AB}^{\prime }{C}^{\prime },{BA}^{\prime }{C}^{\prime }\mathrm{and}{CA}^{\prime }{B}^{\prime }$$$$\mathrm{have}\mathrm{a}\mathrm{common}\mathrm{point}.\mathrm{Prove}\mathrm{that}\mathrm{the}$$$$\mathrm{property}\mathrm{holds}\mathrm{even}\mathrm{if}\mathrm{the}\mathrm{points}{A}^{\prime },$$$$B^{\prime }\mathrm{and}{C}^{\prime }\mathrm{are}\mathrm{collinear}.$$

Answered by mrW1 last updated on 05/Jul/17

Commented by mrW1 last updated on 05/Jul/17

Commented by mrW1 last updated on 05/Jul/17

$${\mathrm{B}}^{\prime }{\mathrm{A}}^{\prime }{\mathrm{C}}^{\prime }\mathrm{are}\mathrm{colinear}.$$$${\mathrm{O}}_{\mathrm{\Delta }{\mathrm{AB}}^{\prime }{\mathrm{C}}^{\prime }}\mathrm{and}{\mathrm{O}}_{\mathrm{\Delta }{\mathrm{CA}}^{\prime }{\mathrm{B}}^{\prime }}\mathrm{intersect}\mathrm{at}\mathrm{M}.$$$$\mathrm{Through}\mathrm{B},{\mathrm{C}}^{\prime },{\mathrm{A}}^{\prime }\mathrm{and}\mathrm{M}\mathrm{one}\mathrm{can}\mathrm{always}$$$$\mathrm{construct}\mathrm{a}\mathrm{circle}.$$$$\Rightarrow \mathrm{all}3\mathrm{circles}\mathrm{have}\mathrm{always}\mathrm{a}\mathrm{common}$$$$\mathrm{point}\mathrm{M},\mathrm{even}\mathrm{more}\mathrm{than}\mathrm{one}.$$

Commented by mrW1 last updated on 05/Jul/17

$${\mathrm{O}}_{\mathrm{\Delta }{\mathrm{AB}}^{\prime }{\mathrm{C}}^{\prime }}\mathrm{and}{\mathrm{O}}_{\mathrm{\Delta }{\mathrm{CA}}^{\prime }{\mathrm{B}}^{\prime }}\mathrm{intersect}\mathrm{at}\mathrm{M}.$$$$\mathrm{\angle }{\mathrm{B}}^{\prime }{\mathrm{MC}}^{\prime }=180°-\mathrm{\angle }\mathrm{A}$$$$\mathrm{\angle }{\mathrm{B}}^{\prime }{\mathrm{MA}}^{\prime }=180°-\mathrm{\angle }\mathrm{C}$$$$\Rightarrow \mathrm{\angle }{\mathrm{B}}^{\prime }{\mathrm{MC}}^{\prime }=360°-\mathrm{\angle }{\mathrm{B}}^{\prime }{\mathrm{MA}}^{\prime }-\mathrm{\angle }{\mathrm{B}}^{\prime }{\mathrm{MC}}^{\prime }$$$$=360°-(180°-\mathrm{\angle }\mathrm{C}+180°-\mathrm{\angle }\mathrm{A})$$$$=\mathrm{\angle }\mathrm{A}+\mathrm{\angle }\mathrm{C}$$$$$$$$\mathrm{\angle }{\mathrm{B}}^{\prime }{\mathrm{MC}}^{\prime }+\mathrm{\angle }\mathrm{B}=\mathrm{\angle }\mathrm{A}+\mathrm{\angle }\mathrm{C}+\mathrm{\angle }\mathrm{B}=180°$$$$\Rightarrow {\mathrm{A}}^{\prime }{\mathrm{BC}}^{\prime }\mathrm{M}\mathrm{is}\mathrm{cyclic}.$$$$\Rightarrow \mathrm{M}\mathrm{lies}\mathrm{on}\mathrm{circle}{\mathrm{O}}_{\mathrm{\Delta }{\mathrm{BA}}^{\prime }{\mathrm{C}}^{\prime }}$$$$$$$$\Rightarrow \mathrm{all}\mathrm{circles}\mathrm{have}\mathrm{a}\mathrm{common}\mathrm{point}\mathrm{M}.$$

Commented by Tinkutara last updated on 05/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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