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

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 if the points A′, B′ and C′ are collinear.

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

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

B′A′C′ are colinear. O_(ΔAB′C′) and O_(ΔCA′B′) intersect at M. Through B,C′,A′ and M one can always construct a circle. ⇒all 3 circles have always a common point M, even more than one.

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

Commented by mrW1 last updated on 05/Jul/17

O_(ΔAB′C′) and O_(ΔCA′B′) intersect at M. ∠B′MC′=180°−∠A ∠B′MA′=180°−∠C ⇒∠B′MC′=360°−∠B′MA′−∠B′MC′ =360°−(180°−∠C+180°−∠A) =∠A+∠C ∠B′MC′+∠B=∠A+∠C+∠B=180° ⇒A′BC′M is cyclic. ⇒M lies on circle O_(ΔBA′C′) ⇒ all circles have a common point M.

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

Commented by Tinkutara last updated on 05/Jul/17

Thanks Sir!

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

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