Let $BE$ and $CF$ be the bisectors of $\angle B$ and $\angle C$ of a triangle $ABC$ whose incentre is $I$. Suppose $EF$, extended, meets the circumcircle of $ABC$ in $M,N$. Show that the circumradius of $MIN$ is twice that of $ABC$.
Problem
Source: Indian Postal Coaching 2007 set 3 p4
Tags: geometry, circumradius, circumcircle, angle bisector
lilavati_2005
26.05.2020 19:17
parmenides51 wrote: Let $BE$ and $CF$ be the bisectors of $\angle B$ and $\angle C$ of a triangle $ABC$ whose incentre is $I$. Suppose $EF$, extended, meets the circumcircle of $ABC$ in $M,N$. Show that the circumradius of $MIN$ is twice that of $ABC$. See here for the same problem and some nice proofs!
jelena_ivanchic
25.02.2021 08:42
I renamed $M,N$ as $P,Q$ ( c'mon $M,N$ are reserved for midpoints ). Also $R(XYZ)$ denotes the radius of circumcircle of $(XYZ)$
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Introduce $BE \cap (ABC)=M_B , CF\cap (ABC) =M_C, I_B,I_C$ as excenters.
Claim: $I_C, I_B \in ( IPQ)$
Proof: Consider POP of $F.$ We get $I_cF \times IF= AF\times FB = FP\times FQ.$
Note that $M_C, M_B$ are midpoints of $II_C, II_B.$ So $R( IPQ)=R(II_CI_B)= 2\times R (IM_CM_B).$
But $M_CM_B||I_CI_B$ and $I_A \perp I_CI_B \implies \Delta AM_CM_B \cong IM_CM_B \implies 2\times R(IM_BM_C)= 2\times R(AM_CM_B)=2\times R(ABC).$
And we are done!