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Question-184041




Question Number 184041 by Acem last updated on 02/Jan/23
Answered by HeferH last updated on 02/Jan/23
TN^(⌢)  = (π/3) = ((180°)/3) = 60°  ⇒ ∠TAN = ((TN^(⌢) )/2) = 30°   ∠AOT = 120° (△AOT isosceles)    ∠ATC = ((AT^(⌢) )/2) = 60  ⇒ ∠ACT = 90°    MTCA is cyclic,   the center would be the mid point of AT
$$\overset{\frown} {{TN}}\:=\:\frac{\pi}{\mathrm{3}}\:=\:\frac{\mathrm{180}°}{\mathrm{3}}\:=\:\mathrm{60}°\:\:\Rightarrow\:\angle{TAN}\:=\:\frac{\overset{\frown} {{TN}}}{\mathrm{2}}\:=\:\mathrm{30}° \\ $$$$\:\angle{AOT}\:=\:\mathrm{120}°\:\left(\bigtriangleup{AOT}\:{isosceles}\right)\: \\ $$$$\:\angle{ATC}\:=\:\frac{\overset{\frown} {{AT}}}{\mathrm{2}}\:=\:\mathrm{60}\:\:\Rightarrow\:\angle{ACT}\:=\:\mathrm{90}°\: \\ $$$$\:{MTCA}\:{is}\:{cyclic}, \\ $$$$\:{the}\:{center}\:{would}\:{be}\:{the}\:{mid}\:{point}\:{of}\:{AT} \\ $$
Answered by mr W last updated on 02/Jan/23
Commented by mr W last updated on 02/Jan/23
∠TAN=((TN^(⌢) )/2)=30°  ∠BOT=2∠OAT=60°=∠OAC  ⇒AC//OT  ⇒AC⊥BC  ⇒∠ACB=90°=∠AMT  ⇒ACTM are cyclic, lie on a circle       with AT as diameter. Midpoint       of AT is the center.
$$\angle{TAN}=\frac{\overset{\frown} {{TN}}}{\mathrm{2}}=\mathrm{30}° \\ $$$$\angle{BOT}=\mathrm{2}\angle{OAT}=\mathrm{60}°=\angle{OAC} \\ $$$$\Rightarrow{AC}//{OT} \\ $$$$\Rightarrow{AC}\bot{BC} \\ $$$$\Rightarrow\angle{ACB}=\mathrm{90}°=\angle{AMT} \\ $$$$\Rightarrow{ACTM}\:{are}\:{cyclic},\:{lie}\:{on}\:{a}\:{circle} \\ $$$$\:\:\:\:\:{with}\:{AT}\:{as}\:{diameter}.\:{Midpoint} \\ $$$$\:\:\:\:\:{of}\:{AT}\:{is}\:{the}\:{center}. \\ $$

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