Let $ABC$ be an isosceles triangle with $AB = AC$. Let $X$ and $K$ points over $AC$ and $AB$, respectively, such that $KX = CX$. Bisector of $\angle AKX$ intersects line $BC$ at $Z$. Show that $XZ$ passes through the midpoint of $BK$.
Problem
Source: 1st Girls in Mathematics Tournament 2019 p5 (Brazil) / Torneio Meninas na Matematica (TM^2 )
Tags: geometry, midpoint, isosceles, equal segments, angle bisector, projective geometry
dgreenb801
26.05.2020 04:50
See my solution on my Youtube channel here: https://www.youtube.com/watch?v=QXus3AwOqW8&t=
Rafinha
13.07.2021 18:41
Projecting by Z we know that $(B,K;M,A)=(C,N;X,A)$=> $\dfrac{MB}{MK} \times \dfrac{AK}{AB}= \dfrac{XC}{XN} \times \dfrac{AN}{AC}$=> as AB=AC and KX=CX we have that $\dfrac{MB}{MK}=\dfrac{XC \times AN}{XN \times AK}$ using the external bisector theorem we have $\dfrac{AK \times XC}{KX \times AK}=\dfrac{MB}{MK}=1$
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PennyLane_31
25.10.2023 04:50
Here’s another solution that a friend shared with me: Define $M= KB\cap XZ$ and $KY$ the bisector of $\angle AKX$. So, $\angle AKY=\angle YKX= \angle ZKB=\alpha$. Also, we know that, by the Law of Sines in the triangles $ZKM, ZKX, ZBM, ZCX$: Law of sines in $ZKM: \frac{sin\angle \alpha}{ZM}= \frac{sin\angle KZM}{KM}\implies \frac{sin\alpha}{sin\angle KZM}= \frac{ZM}{KM}$. (1) Law of sines in $ZKX: \frac{sin\angle ZKX}{ZX}=\frac{sin\angle KZX}{KX}\implies \frac{sin(180-\angle YKX)}{ZX}=\frac{sin\angle KZM}{KX}\implies \frac{sin\alpha}{ZX}=\frac{sin\angle KZM}{KX}$ $\implies \frac{sin\alpha}{sin\angle KZM}= \frac{ZX}{KX}$ (2) Law of sines in $ZBM: \frac{sin\angle MZB}{MB}= \frac{sin\angle MBZ}{ZM}\implies \frac{sin\angle MZB}{MB}= \frac{sin (180-\angle ABC)}{ZM}\implies \frac{sin\angle MZB}{MB}= \frac{sin\angle ABC}{ZM}$ $\implies \frac{sin\angle MZB}{MB}= \frac{sin\angle ACB}{ZM} \implies \frac{sin\angle MZB}{sin\angle ACB}= \frac{MB}{ZM}$(3) Law of sines in $ZCX: \frac{sin\angle XZC}{XC}=\frac{sin\angle XCZ}{XZ}\implies \frac{sin\angle MZB}{XC}= \frac{sin\angle ACB}{XZ} \implies \frac{sin\angle MZB}{sin\angle ACB}= \frac{XC}{XZ}$(4) By (1) and (2): $\frac{ZM}{KM}=\frac{sin\alpha}{sin\angle KZM}= \frac{ZX}{KX}\implies \frac{ZM}{KM}=\frac{ZX}{KX}$ (5) By (3) and (4): $\frac{MB}{ZM}= \frac{sin\angle MZB}{sin\angle ACB}= \frac{XC}{XZ}\implies \frac{MB}{ZM}=\frac{XC}{XZ}$(6) As $KX= XC$, we have: $\frac{ZM}{KM}=\frac{ZX}{XC}$ (by (5)) and $\frac{ZM}{MB}=\frac{ZX}{XC}\implies \frac{ZM}{KM}=\frac{ZX}{XC}= \frac{ZM}{MB}$ $\therefore \frac{\cancel{ZM}}{KM}=\frac{\cancel{ZM}}{MB}\implies KM= MB$, as we wanted
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