In triangle $ABC$, the incircle, with center $I$, touches the sides $BC$ at point $D$. Line $DI$ meets $AC$ at $X$. The tangent line from $X$ to the incircle (different from $AC$) intersects $AB$ at $Y$. If $YI$ and $BC$ intersect at point $Z$, prove that $AB=BZ$. Proposed by Hooman Fattahimoghaddam
Problem
Source: 4th Iranian Geometry Olympiad (Advanced) P1
Tags: geometry, Iran, IGO
15.09.2017 13:33
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 $ZD=AF\implies AB=BZ$.
15.09.2017 21:29
$I$ 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
06.01.2018 16:46
@bgn ,can you post a link to IGO website?
06.01.2018 20:30
liekkas wrote: @bgn ,can you post a link to IGO website? http://igo-official.ir/?lang=en
06.01.2018 20:54
Also, the problems of 2017 IGO: https://artofproblemsolving.com/community/c520386_2017_iranian_geometry_olympiad
07.01.2018 10:26
Thanks above
20.08.2018 11:26
03.09.2018 14:02
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.
03.09.2018 15:56
http://mathworld.wolfram.com/notebooks/PlaneGeometry/NeubergCubic.nb
30.05.2020 11:58
Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/26.%203.%20Segments.pdf p. 9-10. Sincerely Jean-Louis
25.11.2020 01:32
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 $AB=BZ$
03.07.2021 10:50
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 $BI$.
03.11.2021 16:38
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!!!
03.11.2021 17:11
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.
27.12.2021 16:07
∠XIY = 90 - ∠A/2 ---> ∠IZB = ∠A/2 now triangles IZB and IAB are congruent so AB = BZ.
12.05.2022 03:24
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 $\triangle ABI\cong\triangle ZBI.$ $\square$
06.02.2024 16:24
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.
02.07.2024 13:01
Angle chase and some isométrique triangle