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Question-216958

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Question Number 216958 by abdi last updated on 25/Feb/25
Answered by A5T last updated on 25/Feb/25
Commented by A5T last updated on 26/Feb/25
2) Let circle passing through CHB and ACB be Ω and Γ respectively ∠CHB=90° ⇒CB is the diameter of Ω ∠ACB=90°⇒AC is tangent to Ω Power of a point with respect to Ω from A ⇒AC^2 =AH×AB. ∠ACB=90°⇒ AB is the diameter of Γ ∠CHB=90°⇒AB bisects CD⇒CH=HD Power of a point with respect to Γ from H ⇒AH×BH=CH×HD=CH^2
$$\left.\mathrm{2}\right) \\ $$$$\mathrm{Let}\:\mathrm{circle}\:\mathrm{passing}\:\mathrm{through}\:\mathrm{CHB}\:\mathrm{and}\:\mathrm{ACB}\:\mathrm{be}\: \\ $$$$\Omega\:\mathrm{and}\:\Gamma\:\mathrm{respectively} \\ $$$$\angle\mathrm{CHB}=\mathrm{90}°\:\Rightarrow\mathrm{CB}\:\mathrm{is}\:\mathrm{the}\:\mathrm{diameter}\:\mathrm{of}\:\Omega \\ $$$$\angle\mathrm{ACB}=\mathrm{90}°\Rightarrow\mathrm{AC}\:\mathrm{is}\:\mathrm{tangent}\:\mathrm{to}\:\Omega \\ $$$$\mathrm{Power}\:\mathrm{of}\:\mathrm{a}\:\mathrm{point}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\Omega\:\mathrm{from}\:\mathrm{A} \\ $$$$\Rightarrow\mathrm{AC}^{\mathrm{2}} =\mathrm{AH}×\mathrm{AB}. \\ $$$$\angle\mathrm{ACB}=\mathrm{90}°\Rightarrow\:\mathrm{AB}\:\mathrm{is}\:\mathrm{the}\:\mathrm{diameter}\:\mathrm{of}\:\Gamma \\ $$$$\angle\mathrm{CHB}=\mathrm{90}°\Rightarrow\mathrm{AB}\:\mathrm{bisects}\:\mathrm{CD}\Rightarrow\mathrm{CH}=\mathrm{HD} \\ $$$$\mathrm{Power}\:\mathrm{of}\:\mathrm{a}\:\mathrm{point}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\Gamma\:\mathrm{from}\:\mathrm{H} \\ $$$$\Rightarrow\mathrm{AH}×\mathrm{BH}=\mathrm{CH}×\mathrm{HD}=\mathrm{CH}^{\mathrm{2}} \\ $$
Answered by A5T last updated on 25/Feb/25
Commented by A5T last updated on 25/Feb/25
3) Van Aubel′s theorem : ((AG)/(GF))=((AD)/(DB))+((AE)/(EC)) ⇒((AG)/(GF))=(1/1)+(1/1)=2⇒AG=2GF Similarly, we get BG=2GE ; CG=2DG
$$\left.\mathrm{3}\right) \\ $$$$\mathrm{Van}\:\mathrm{Aubel}'\mathrm{s}\:\mathrm{theorem}\::\:\frac{\mathrm{AG}}{\mathrm{GF}}=\frac{\mathrm{AD}}{\mathrm{DB}}+\frac{\mathrm{AE}}{\mathrm{EC}} \\ $$$$\Rightarrow\frac{\mathrm{AG}}{\mathrm{GF}}=\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}}=\mathrm{2}\Rightarrow\mathrm{AG}=\mathrm{2GF} \\ $$$$\mathrm{Similarly},\:\mathrm{we}\:\mathrm{get}\:\mathrm{BG}=\mathrm{2GE}\:;\:\mathrm{CG}=\mathrm{2DG} \\ $$
Answered by A5T last updated on 25/Feb/25
Commented by A5T last updated on 25/Feb/25
Menelaus′ Theorem: ((AC_1 )/(C_1 B))×((BC)/(CA_1 ))×((A_1 D)/(DA))=1 ⇒(1/1)×(3/1)×((A_1 D)/(DA))=1⇒AD:DA_1 =3:1
$$\mathrm{Menelaus}'\:\mathrm{Theorem}:\:\frac{\mathrm{AC}_{\mathrm{1}} }{\mathrm{C}_{\mathrm{1}} \mathrm{B}}×\frac{\mathrm{BC}}{\mathrm{CA}_{\mathrm{1}} }×\frac{\mathrm{A}_{\mathrm{1}} \mathrm{D}}{\mathrm{DA}}=\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{1}}×\frac{\mathrm{3}}{\mathrm{1}}×\frac{\mathrm{A}_{\mathrm{1}} \mathrm{D}}{\mathrm{DA}}=\mathrm{1}\Rightarrow\mathrm{AD}:\mathrm{DA}_{\mathrm{1}} =\mathrm{3}:\mathrm{1} \\ $$

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