The incircle $\omega$ of a triangle $ABC$ touches $BC, AC$ and $AB$ at points $A_0, B_0$ and $C_0$ respectively. The bisectors of angles $B$ and $C$ meet the perpendicular bisector to segment $AA_0$ at points $Q$ and $P$ respectively. Prove that $PC_0$ and $QB_0$ meet on $\omega$ .
I think I have surely made some mistake,because this seems little easy for 10-11(Can someone please check this?)
Solution:Clearly $ABA_0Q$ is cyclic with $Q$ as the midpoint of arc $AA_0$ not containing $B$.Similarly $P$.$\angle PQA_0=90-\angle AA_0Q=90-\frac {B}{2}$
Similarly $\angle QPA=90-\frac {C}{2}$.Now by spiral similarity on $\triangle A_0B_0C_0$ and $\triangle PA_0Q$ the conclusion follows
mmathss wrote:
I think I have surely made some mistake,because this seems little easy for 10-11(Can someone please check this?)
Solution:Clearly $ABA_0Q$ is cyclic with $Q$ as the midpoint of arc $AA_0$ not containing $B$.Similarly $P$.$\angle PQA_0=90-\angle AA_0Q=90-\frac {B}{2}$
Similarly $\angle QPA=90-\frac {C}{2}$.Now by spiral similarity on $\triangle A_0B_0C_0$ and $\triangle PA_0Q$ the conclusion follows
I think you are right