WLOG let $AC\ge AB$. Note that $\angle{DXC}=90-\angle{C}\implies \angle{YXD}=90-\angle{C}$. So from quadrilateral $BYXD$ we get $$\angle{BYZ}=\angle{ZYX}=90+\dfrac{\angle{C}-\angle{B}}{2}.$$Thus we have $\angle{YZB}=\angle{A}/2$. Hence $\triangle{AIF}\cong \triangle{ZID}$ implying that $ZD=AF\implies AB=BZ$.

$I$ is the excenter of $\triangle ABC$, thus $\angle BYI=\widehat{C}+\frac{\hat A}2$, hence $\angle BZI=\frac{\hat A}2=\angle BAI$, done. Best regards, sunken rock

@bgn ,can you post a link to IGO website?

liekkas wrote: @bgn ,can you post a link to IGO website? http://igo-official.ir/?lang=en

Also, the problems of 2017 IGO: https://artofproblemsolving.com/community/c520386_2017_iranian_geometry_olympiad

Thanks above

Can someone explain this: Solution. In triangle AXY , the point I is excenter so <XIY=90-<A/2 It s a part of the solution IGO published.

http://mathworld.wolfram.com/notebooks/PlaneGeometry/NeubergCubic.nb

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/26.%203.%20Segments.pdf p. 9-10. Sincerely Jean-Louis

Our end-goal is to prove that $DF \parallel AZ$ Let $T$ be the second tangent from $X$ onto the incircle. Throw the configuration onto the complex plane, and let every point in the Euclidean plane be represented by its' lowercase letter and the incircle is the unit circle. We claim W.L.O.G. that $d=1$. Then we have that: $$a = \frac{2ef}{e+f}$$$$b=\frac{2f}{f+1}$$$$c=\frac{2e}{e+1}$$ We have that $x$ belongs to the real number line, thus we have that $x = \overline{x}$ Since we have that $\frac{e-a}{\overline{e-a}}=\frac{e-x}{\overline{e-x}}$, this implies that $x=\frac{2e}{1+e^2}$ By definition and from $x$, we have that $t=\frac{1}{e}$, this implies that $y=\frac{2f}{ef+1}$. Since we have that $z$ must satisfy these conditions, $\frac{y}{\overline{y}}=\frac{y-z}{\overline{y-z}}$ and $\frac{b-c}{\overline{b-c}}=\frac{b-z}{\overline{b-z}}$, we have that $z=\frac{2f}{f+e}$ By easy computation we see that: $$\frac{z-a}{\overline{z-a}}=\frac{d-f}{\overline{d-f}}$$this implies that $DF \parallel AZ$, from which we have that $AB=BZ$

Let $T$ be the point on $AC$ such that $ZT$ is tangent to the incircle. By Brianchon's, $T$ is on $BI$, so $A$ and $Z$ are symmetric about $BI$.

Drop a tangent from $Z$ to the incircle and let it meet $AC$ at $E$. By biranchon on the degenerate hexagon $BYXEZD$ we have $E,I,B$ colinear and since $\angle AEB=\angle ZEB$ we have $A,Z$ symetric w.r.t. $BI$ thus $AB=BZ$ as desired.

Edit: 600th post yay!!!

This is very easy. The quadrilateral $BYXC$ has an incircle and the angle bisectors concur at $I$.Some trivial angle chasing (or angle halfing and doubling) gives that $\angle IZB=\frac{A}{2} \implies AIZC$ is cyclic .So $\angle AZC=90+\frac{B}{2}$, as desired.

∠XIY = 90 - ∠A/2 ---> ∠IZB = ∠A/2 now triangles IZB and IAB are congruent so AB = BZ.

Notice $$\angle IZB=90-\angle DIZ=\angle IZX+\angle YXI-90=\tfrac{1}{2}(\angle BYX+\angle YXC-180)=\tfrac{1}{2}\angle A$$so $\triangle ABI\cong\triangle ZBI.$ $\square$

Let $E$ be the incenter of $\Delta AXY$ and $F$ be the tangency point of $AB$ and the incircle of $\Delta ABC$. The incenter of $\Delta ABC$ is the $A$-excircle of $\triangle AXY$. So, by Incenter-Excenter Lemma, $EXIY$ is cyclic. \[\angle AIY = \angle EIY = \angle EXY=\frac{1}{2}\angle AXY = \frac{1}{2}(180^{\circ}-2\angle CXD) = \frac{1}{2}(180^{\circ}-(180^{\circ}-2\angle C)) = \frac{1}{2}\cdot 2\angle C=\angle ACZ\implies AIZC\text{ is cyclic}\]Finally, \[\angle AZI = \angle ACI=\angle ZCI = \angle ZAI\implies IA=IZ\implies \triangle AIF\cong\triangle IDZ\implies AF=DZ\implies AF+BF=BD+DZ\implies AB=BZ\]as desired.

Angle chase and some isométrique triangle