As i promised before , here i am going to give my solution for
Problem 38 .Solution of Problem 38 :Lemma 1: In a triangle $\triangle ABC$ ,let $M$ be the midpoint of $BC$ .Let $D$ be the projection of $A$ on $BC$ .Let $E$ be a point such that ,$E$ and $A$ are on the same side of $BC$ , $ED \| AM$ and $BM^2=AM( 2ED +AM)$ . Then $\angle BAC =180^\circ -\dfrac{1}{2}\angle BEC$
Proof : Let $K , L$ be the reflection of $A$ wrt $BC$ and $M$ respectively . Then $BKLC$ is cyclic .Let $AL $ intersect $\odot BKLC$ again at $J$ .Then , $$ ML \cdot JM =BM\cdot MC=BC^2=AM( 2ED +AM)=ML( 2ED +AM)$$
$$\Rightarrow JM = 2ED +AM \Rightarrow JA = 2ED $$
Now , as $JA \| ED$ and $D$ is the midpoint of $AK$ , So $J ,E ,K$ are collinear and $E$ is the midpoint of $JK$ .Let $O$ be the center of $\odot BKLC$ . Then $OE \perp EK$ . Now as , $arc ~BK =arc ~LC$ .So , $JK$ is the $J$ -symmedian of $\triangle JBC$ . So tangents to $\odot BKLC$ at $B$ and $C$ meet at $S \in JK$ .Then $BEOCS
$ is cyclic .
So , $$\angle BAC=\angle BLC =180^\circ -\angle BJC =180^\circ -\dfrac{1}{2}\angle BOC =180^\circ -\dfrac{1}{2}\angle BEC$$
Now we will use this lemma to prove a beautiful result on angles related to some famous triangle centers .
A Beautiful Lemma : Let $O$ and $N$ be the circumcenter and nine point center of $\triangle ABC$ repectively . Let $I_a ,I_b$ and $I_c$ be $A$-excenter , $B$-excenter and $C$-excenter of $\triangle ABC$ repectively . Then , $\angle I_bOI_c=180^\circ -\dfrac{1}{2}\angle I_bNI_c $
Comment:Proof :Let the circumcircle of $\triangle ABC $ meet $I_cI_b$ again at $M$. Then $M$ is the midpoint of $I_cI_b$ . Let $L$ be the midpoint of $AM$ , then $OL \perp I_cI_b$ and $LN \perp BC$ . Now $LN \| MO$ as $MO \perp BC$ .Let $MO$ intersect $BC$ at $P$ and $\odot ABC$ again at $S$ .Let $K$ be the midpoint of $MO$ . Then $LK=\dfrac{R}{2}=NP$ , so $LKPN$ is a parallelogram .So $LN=KP$ .Now ,
$$I_bM^2=MC^2=MS\cdot MP=2OM(MK+KP)=2OM(\dfrac{OM}{2} + LN)=OM(2LN + OM)$$
Now the proof is done , by using lemma 1 .
Lemma 3: In $\triangle ABC$ , $I_aO \perp B_oC_o$
Proof : Let $T$ be the circumcenter of $\triangle I_cII_b$ , then $T,O,I_a$ are collinear as $O$ and $I_a$ are the ninepoint center and orthocenter of $\triangle I_cII_b$ respectively . $AIBI_c$ and $AICI_b$ are cyclic . So , $AC_o \cdot BC_o=IC_o \cdot I_cC_o $ and $AB_o \cdot CB_o=IB_o \cdot I_bB_o $. So $ B_oC_o$ is the radical axis of $\odot I_cII_b$ and $\odot ABC$ . So , $I_aO \perp B_oC_o$.
Lemma 4: Let the $A$-excircle , $B$-excircle and $C$-excircle of $\triangle ABC$ touch the ninepoint circle at $L_a , L_b $ and $L_c$ repectively . Then $\triangle L_aL_bL_c \sim \triangle A_oB_oC_o$
Proof : By using lemma 2 we get ,
$$\angle L_aL_bL_c =\dfrac{1}{2} \angle L_aNL_c =\dfrac{1}{2} \angle I_aNI_c =180^\circ - \angle I_bOI_c $$
Using lemma 3 we get , $$\angle A_oB_oC_o = 180^\circ - \angle I_bOI_c $$
So , $$\angle A_oB_oC_o = \angle L_aL_bL_c $$
Similarly , we can show this for other angles .So , $\triangle L_aL_bL_c \sim \triangle A_oB_oC_o$
Lemma 5: Let $F_e$ be the feuerbach point of $\triangle ABC$. Then $F_e , A_o$ and $L_a$ are colinear .
Proof: Let $M_a$ be the midpoint of $BC$ .$H_a$ be the projection of $A$ on $BC$ .The incircle and $A$ -excircle touches $BC$ at $X,Y$ respectively .Consider the inversion with center $M_a$ and radius $M_aY$.For a point $W$ , let $\bar W$ denote its inverse .$(H_a,A_o;X,Y)=-1$. So $H_a$ is the inverse of $A_o$ . Now , $\bar F_e\bar L_a$ is the other internal common tangent of the incircle and $A$ -excircle. $\bar L_aA_o \cdot A_o\bar F_e=XA_o \cdot A_oY=H_aA_o \cdot A_oM_a $ . So , $M_a\bar F_eH_a\bar L_a$ is cyclic . So , $F_e , A_o$ and $L_a$ are colinear .
Back To The Main Problem : $\measuredangle B_oF_eC_o =\measuredangle L_bF_eL_c=\measuredangle L_bL_aL_c= \measuredangle B_oA_oC_o$ . So $A_o , B_o , C_o , F_e$ are concyclic .