Geometry

Two intersecting circles $ \displaystyle {{C}_{1}}\ \text{and  }{{C}_{2}}$ have a common tangent which touches $ \displaystyle {{C}_{1}}\ \text{at  }P\ \ \text{and  }{{C}_{2}}\ \ \text{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}}$ again at R . Prove that MQ bisects $ \displaystyle \angle PMR$ .

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

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

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

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

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \text{In }\Delta \ PQR,\ \ \ \theta +\phi =\Omega $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \beta +\gamma =\Omega $

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

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

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \beta +\gamma =\alpha $

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