I think that something is not correct, in the notation of this proposition. In my drawing, BC = 13.0, BD = 2.6, ID = 1.8, where I, is the incenter of ABC and D, the touch point of the incircle to the sideline BC. So, I see that the lines IQ, AK, intersect each other and the triangles AIG, QPG, have not equal areas. Sorry, if I am mistaken. Kostas Vittas.

I think the point K is defined incorrectly, it should be simply CK = BD, not CK = 3 BD. Then K, L are tangency points of BC, CA with excircles against A, B, respectively, and AK, BL are perimeter splitters intersecting at the Nagel point P, so that I, G, P are collinear and GP = 2 GI. Since GA = 2 GQ and the vertical angles $\angle AGI = \angle QGP$ are equal, ${|\triangle AIG| = \frac{1}{2}GA \cdot GI \sin \widehat{AGI}= \frac{1}{2}GQ \cdot GB \sin \widehat{QGP}}= |\triangle QPG|$ Q is the midpoint of both BC and KD. QI cuts AD at its midpoint M, because the Newton line QM of degenerate tangential quadrilateral ABDC with diagonals AD, BC passes through its incenter I. Therefore, QM is the midline of $\triangle ADK$ parallel to AK.

I think that this is a good correction yetti. Although it is very nice your approach by Newton’s theorem (useful for somebody who is interested to learn new thinks in geometry ), we can easy to prove that IQ // AK, because of GA / GQ = GP / GI = 2. From trapezium AIQP now, we have that (AIP) = (AQP) => (AIG) + (AGP) = (QPG) + (AGP) => (AIG) = (QPG). Kostas Vittas.

a solution for part A: We assume that $AC=b \geq c=AB.$We need to prove $\frac{NI}{IA}=\frac{NQ}{QK}$.We know that $BN=\frac{ac}{b+c}$ and $CN=\frac{bc}{b+c}$.Since $BI$ is angle bisector, $\frac{NI}{IA}=\frac{BN}{AB}=\frac{{(\frac{ac}{b+c})}}{c}=\frac{a}{b+c}....(1)$ $NQ=BQ-BN=\frac{a}{2}-\frac{ac}{b+c}=\frac{ab+ac-2ac}{2(b+c)}=\frac{a(b-c)}{2(b+c)}....(2)$ $QK=QC-KC=\frac{a}{2}-(u-b)=\frac{a-2u-2b}{2}=\frac{b-c}{2}....(3)$ from $(2)$ and $(3)$, $\frac{NQ}{QK}=\frac{a(b-c)}{2(b+c)}.\frac{2}{b-c}=\frac{a}{b+c}....(4)$ from $(1)$ and $(4)$, $\frac{NI}{IA}=\frac{NQ}{QK}$ and $IQ \parallel AK$ Lokman Gökçe

Attachments:

Fistly，we assume that ⊙J is the excircle of triangle ABC the center J of which is on the bisector of ∠A,and ⊙J touches AB at X, BC at K' and CA at Y. ⊙I touches AB at Z. Secondly,we assmue AZ=AE=x,BZ=BD=y,CD=CE=u,BK'=BX=v,CK'=CY=w,and then we know that AX=AY and BC=BK+KC=BK'+K'C. These mean that x+y+v=x+u+w and y+u=v+w,so we can find that y=w.At last we can draw a conclusion K and K' are the same point. Extend DI to meet ⊙I at K". Please pay attention to the fact that ⊙I and ⊙J are homothetic figures, so we can be sure of that K" is also on AK, so DI/IK"=DQ/QK=1, so IQ//AK!

Your solution nice, Butterfly.I liked. I'm posting my other solution: Let be $IQ \bigcap AD = \{M\}$.Clearly, $Q$ is mid-point of $[KD]$.If we can prove $AM=MD$ then, $IQ//AK$ showed. $\frac{QN}{ND}\cdot\frac{DM}{MA}\cdot\frac{AI}{IN}=1...(*)$ by Menelaus and together $\frac{AI}{IN}=\frac{b+c}{a}....(1)$ $QD=QK=\frac{b-c}{2}....(**)$ $QN=BQ-BN=\frac{a}{2}-\frac{ac}{b+c}=\frac{ab+ac-2ac}{2(b+c)}=\frac{a(b-c)}{2(b+c)}....(***)$ from $(**)$ and $(***)$, $\frac{QN}{QD}=\frac{a(b-c)}{2(b+c)}\cdot\frac{2}{(b-c)}=\frac{a}{b+c}....(****)$ from $(1),(*),(****)$ , $DM=MA$.Hence, $IQ//AK$. Lokman GÖKÇE

Your solution also wonderful Mr.Lokman ,thank you.We would be very grateful to see you here.

Another solution: Using the second lemma from this wonderful link http://web.mit.edu/yufeiz/www/olympiad/geolemmas.pdf we have $P$ is the Nagel Point of ABC and by Nagel line we know that $I,G,Na(P)$ points are collinear. Then again using a lemma $2IG=GNa(P)$ hence ; $\Delta_{ABC}=\frac{sinAGI.AG.GI}{2}=\frac{sinPGQ.PG.QG}{2}=\Delta_{PGQ}$ hence we are done