euclidean-geometrygeometry
In $\triangle ABC$ it holds $BC = \frac{AB+AC}{2}$ .Let $M$ and $N$ be midpoints of $AB$ and $AC$ ,and let $I$ be the incentre of $\triangle ABC$. Prove that $A,M,I,N$ are concyclic.
I have tried it very hard. But there are some things that i could decipher.
$$\frac{AO}{OH}=\frac{AM}{MB}=\frac{AN}{NC}=1$$. Rest are drawn on the figure.
Let the midpoints of $CB$ and $CA$ be $A'$ and $B'$. Now we place a point on $D$ on $AB$ such that $AD=AB'$. And as $2c=a+b$, we also have $BD=BA'$.
Thus, by SAS congruence, $\triangle BIA'\cong \triangle BID$ and $\triangle AIB'\cong \triangle AID$. Thus, $\angle BDI= \angle IA'B$ and $\angle ADI =\angle IB'A$. Hence, $\angle IB'A=180^{\circ}-\angle IA'B=\angle IA'C$.
Thus, $C$, $B'$, $A'$ and $I$ lie on a circle.
$AHC$ is a right triangle and the median relative to hypotenuse is half the hypotenuse
$AH=\frac12 FD\to AH=AD$
$ADCB$ is a parallelogram so AD=BC$
therefore $AH=BC$
Furthermore $AB\parallel DE$ and this proves that $ABCH$ is an isosceles trapezoid which is always possible to inscribe in a circle.
Hope this helps
EDIT
The median $AH$ is half $DF$ because you can build a rectangle as I did in the second picture, diagonals are equal and bisect each other. $$...$$
Best Answer
hint:
connect $BI,CI$, check $\triangle MIB$ and $\triangle HIB$,what is their relation. then what is the relation between $\angle IMB$ and $\angle IHB$.
similar method, check $\angle INC$ and $\angle IHC $,now you can prove the result.