Menu Close

Question-91836




Question Number 91836 by  M±th+et+s last updated on 03/May/20
Commented by  M±th+et+s last updated on 03/May/20
let A B C D be four points on a circle  let four more cirvces pass through   AB , BC , CD , DA Respectively,  meeting infurther points P,Q,R,S.  show that P,Q,R and S are consyclic.
$${let}\:{A}\:{B}\:{C}\:{D}\:{be}\:{four}\:{points}\:{on}\:{a}\:{circle} \\ $$$${let}\:{four}\:{more}\:{cirvces}\:{pass}\:{through}\: \\ $$$${AB}\:,\:{BC}\:,\:{CD}\:,\:{DA}\:{Respectively}, \\ $$$${meeting}\:{infurther}\:{points}\:{P},{Q},{R},{S}. \\ $$$${show}\:{that}\:{P},{Q},{R}\:{and}\:{S}\:{are}\:{consyclic}. \\ $$
Answered by mr W last updated on 04/May/20
Commented by mr W last updated on 04/May/20
Commented by mr W last updated on 07/May/20
α_1 +α_4 =α_2 +α_3 =π, since ABQP is cyclic.  β_1 +β_4 =β_2 +β_3   γ_1 +γ_4 =γ_2 +γ_3   δ_1 +δ_4 =δ_2 +δ_3   (α_1 +δ_2 )+(α_4 +δ_3 )=(α_2 +δ_1 )+(α_3 +δ_4 )  (β_2 +γ_1 )+(β_3 +γ_4 )=(β_1 +γ_2 )+(β_4 +γ_3 )  (α_1 +δ_2 +β_2 +γ_1 )+(α_4 +δ_3 +β_3 +γ_4 )=(γ_2 +δ_1 +α_2 +β_1 )+(α_3 +δ_4 +β_4 +γ_3 )  α_4 +β_3 +γ_4 +δ_3 =α_3 +δ_4 +β_4 +γ_3   2π−θ_B +2π−θ_D =2π−θ_A +2π−θ_C   ⇒θ_B +θ_D =θ_A +θ_C   since θ_A +θ_B +θ_C +θ_D =2π  ⇒θ_B +θ_D =θ_A +θ_C =π  i.e. PQRS is cyclic.
$$\alpha_{\mathrm{1}} +\alpha_{\mathrm{4}} =\alpha_{\mathrm{2}} +\alpha_{\mathrm{3}} =\pi,\:{since}\:{ABQP}\:{is}\:{cyclic}. \\ $$$$\beta_{\mathrm{1}} +\beta_{\mathrm{4}} =\beta_{\mathrm{2}} +\beta_{\mathrm{3}} \\ $$$$\gamma_{\mathrm{1}} +\gamma_{\mathrm{4}} =\gamma_{\mathrm{2}} +\gamma_{\mathrm{3}} \\ $$$$\delta_{\mathrm{1}} +\delta_{\mathrm{4}} =\delta_{\mathrm{2}} +\delta_{\mathrm{3}} \\ $$$$\left(\alpha_{\mathrm{1}} +\delta_{\mathrm{2}} \right)+\left(\alpha_{\mathrm{4}} +\delta_{\mathrm{3}} \right)=\left(\alpha_{\mathrm{2}} +\delta_{\mathrm{1}} \right)+\left(\alpha_{\mathrm{3}} +\delta_{\mathrm{4}} \right) \\ $$$$\left(\beta_{\mathrm{2}} +\gamma_{\mathrm{1}} \right)+\left(\beta_{\mathrm{3}} +\gamma_{\mathrm{4}} \right)=\left(\beta_{\mathrm{1}} +\gamma_{\mathrm{2}} \right)+\left(\beta_{\mathrm{4}} +\gamma_{\mathrm{3}} \right) \\ $$$$\left(\alpha_{\mathrm{1}} +\delta_{\mathrm{2}} +\beta_{\mathrm{2}} +\gamma_{\mathrm{1}} \right)+\left(\alpha_{\mathrm{4}} +\delta_{\mathrm{3}} +\beta_{\mathrm{3}} +\gamma_{\mathrm{4}} \right)=\left(\gamma_{\mathrm{2}} +\delta_{\mathrm{1}} +\alpha_{\mathrm{2}} +\beta_{\mathrm{1}} \right)+\left(\alpha_{\mathrm{3}} +\delta_{\mathrm{4}} +\beta_{\mathrm{4}} +\gamma_{\mathrm{3}} \right) \\ $$$$\alpha_{\mathrm{4}} +\beta_{\mathrm{3}} +\gamma_{\mathrm{4}} +\delta_{\mathrm{3}} =\alpha_{\mathrm{3}} +\delta_{\mathrm{4}} +\beta_{\mathrm{4}} +\gamma_{\mathrm{3}} \\ $$$$\mathrm{2}\pi−\theta_{{B}} +\mathrm{2}\pi−\theta_{{D}} =\mathrm{2}\pi−\theta_{{A}} +\mathrm{2}\pi−\theta_{{C}} \\ $$$$\Rightarrow\theta_{{B}} +\theta_{{D}} =\theta_{{A}} +\theta_{{C}} \\ $$$${since}\:\theta_{{A}} +\theta_{{B}} +\theta_{{C}} +\theta_{{D}} =\mathrm{2}\pi \\ $$$$\Rightarrow\theta_{{B}} +\theta_{{D}} =\theta_{{A}} +\theta_{{C}} =\pi \\ $$$${i}.{e}.\:{PQRS}\:{is}\:{cyclic}. \\ $$
Commented by  M±th+et+s last updated on 04/May/20
great work sir
$${great}\:{work}\:{sir} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *