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Tuesday, December 11, 2018

Geometry

Two intersecting circles \displaystyle {{C}_{1}}\ \text{and  }{{C}_{2}}C1 and C2 have a common tangent which touches \displaystyle {{C}_{1}}\ \text{at  }P\ \ \text{and  }{{C}_{2}}\ \ \text{at  }QC1 at P  and C2  at Q. The two circles intersect at M and N , where N is nearer to PQ then M . The line PN meets the circle \displaystyle \text{ }{{C}_{2}} C2 again at R . Prove that MQ bisects \displaystyle \angle PMRPMR .



\displaystyle \text{Proof : Join  }MN\text{  and   }QR.Proof : Join MN and QR.

\displaystyle \ \ \ \ \ \ \ \ \ \ \theta =\beta \ \ (\text{angle between tangent and chord = }          θ=β  (angle between tangent and chord = 

\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{angle in the alternate segment )}                      angle in the alternate segment )

\displaystyle \ \ \ \ \ \ \ \ \ \ \phi =\gamma \ \ \ \ \text{( same arc }QN\ )          ϕ=γ    ( same arc QN )

\displaystyle \ \ \ \ \ \ \ \ \ \ \text{In }\Delta \ PQR,\ \ \ \theta +\phi =\Omega           In Δ PQR,   θ+ϕ=Ω

\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \beta +\gamma =\Omega                            β+γ=Ω

\displaystyle \ \ \ \ \ \ \ \ \ \ \Omega =\alpha \text{ }\ (\text{angle between tangent and chord = }          Ω=α  (angle between tangent and chord = 

\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{angle in the alternate segment )}                      angle in the alternate segment )

\displaystyle \ \ \ \ \ \ \ \ \ \ \beta +\gamma =\alpha           β+γ=α

\displaystyle \ \ \ \ \ \ \ \ \ \therefore \ MQ\ \ \text{bisects   }\angle PMR.


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