Let $ABC$ be a triangle with $AB<AC$. Let $D$ be the point where the bisector of angle $\angle BAC$ touches $BC$ and let $D'$ be the reflection of $D$ in the midpoint of $BC$. Let $X$ be the intersection of the bisector of angle $\angle BAC$ with the line parallel to $AB$ that passes through $D'$. Prove that the line $AC$ is tangent with the circumscribed circle of triangle $XCD'$
Problem
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Tags: geometry
rafaello
27.02.2021 21:53
Let $F=AC\cap D'X$ Claim 1. By parallel lines, $\frac{FD'}{FC}=\frac{AB}{AC}$. Claim 2. By the angle bisector theorem, $\frac{FC}{FA}=\frac{CD'}{D'B}=\frac{BD}{CD}=\frac{AB}{AC}$. $$\angle XAF=\angle BAX=\angle FXA\implies FX=FA.$$Hence, we need to show that $FC^2=FA\cdot FD'\Longleftrightarrow \frac{FC}{FA}=\frac{FD'}{FC}$. By the claims $1$ and $2$, we are done.
Circumcircle
27.02.2021 21:57
In fact, the sufficient condition for the problem to hold is $AB \neq AC$, but in the other case the point $X$ is located in the inside of the triangle.
Circumcircle
27.02.2021 22:03
Clearly, $CD'=BD$. Let $N$ be the intersection of the perpendicular bisector of side $BC$ and the angle bisector of $BAC$. It is known that $N$ belongs to $(ABC)$. Since $D'X \parallel AB$ we have that $\angle D'XN = \angle D'XA = \angle BAX = \angle BAN = \angle BCN$. This gives us that $CD'NX$ is cyclic. Since $NB=NC$, $BD=CD'$ and $\angle NBD = \angle NCD'$ we have that $\bigtriangleup BND \cong \bigtriangleup CND'$, thus $\angle BND = \angle CND'$. Now we have all the neccessary relations to finish the solution, $\angle ACB = \angle BNA = \angle BND = \angle CND' = \angle CXD'$. This means that $AC$ is tangent to $(XCD')$.