Prove line connecting intersection of tangents and opposite vertex bisects segment containing intersection of tangents and a vertex

Question

Let $\triangle ABC$ be an isosceles triangle with $AB=BC$. Let $\Gamma$ be the circumcircle of $\triangle ABC$. Let the tangents at $A$ and $B$ intersect at $D$, and let $DC\cap\Gamma=E\neq C$. Prove that $AE$ bisects segment $BD$.

This is my progress: I let $\Gamma_1$ be the circumcircle of $\triangle ADE$, so $AE$ becomes the radical axis of $\Gamma_1,\Gamma$. Now, if I can prove that $\Gamma_1$ is tangent to $BD$ at $D$, I would be done by applying the following well known fact: The radical axis of two intersecting circles (the common chord of the two circles) bisects the common external tangent of the two circles. However I'm having some trouble proving that the circle is tangent at $D$. I would appreciate any ideas on how to do this.

Answer 1

I have a nice solution with complex numbers.

We are free to assume $B=1,A=e^{-i\theta},C=e^{i\theta}$.

Then $D$ is the inverse of the midpoint of $AB$ with respect to the unit circle, hence: $$ D=\frac{2}{1+e^{i\theta}}=\frac{2}{1+C} $$ and by solving $\left\|\lambda C + (1-\lambda)\frac{2}{1+C}\right\|_{2}^{2}=1$ we get $\lambda=1$ or $\lambda=\frac{1}{5+4\cos\theta}$, from which: $$ E = \frac{1}{2}+\frac{3}{2+4C}=\frac{2+4A}{2+4C}\cdot C. $$ Then we have just to check that $A,E$ and the midpoint $M$ of $BD$ are collinear: simple algebra.

Answer 2

Let $|BA|=|BC|=a$, $|AC|=b$, $\angle ABC=\beta$, $\angle CAB=\angle ACB=\alpha=\tfrac\pi2-\tfrac\beta2$.

Since $\triangle AEC \cong\triangle FED$, let's find the scaling factor,

\begin{align} \frac{|EC|}{|DE|} &= \frac{|DC|-|DE|}{|DE|} = \frac{|DC|}{|DE|}-1. \end{align}

From $\triangle DBC$: \begin{align} |DC|^2 &= |BD|^2+a^2-2a|BD|\cos(\tfrac\pi2+\tfrac\beta2) = |BD|^2+a^2+2a|BD|\sin\tfrac\beta2 \end{align}

Also, using the tangent-secant theorem, \begin{align} |DE| \cdot|DC| &= |AD|^2=|BD|^2, \end{align} hence \begin{align} \frac{|DC|}{|DE|} &= 1+ \left(\frac{2a}{|BD|}\right)^2 + \frac{a}{|BD|}\sin\tfrac\beta2 \\ \frac{|EC|}{|DE|} &= \left(\frac{2a}{|BD|}\right)^2 + \frac{a}{|BD|}\sin\tfrac\beta2 \quad (1) \end{align} Substitution \begin{align} \frac{a}{|BD|} &= 2\sin\tfrac\beta2 \end{align} into (1) gives

\begin{align} \frac{|EC|}{|DE|} &= 8\sin^2\tfrac\beta2 \end{align}

Then \begin{align} |FD| &= \frac{|DE|}{|EC|}\cdot b = \frac{b}{8\sin^2\tfrac\beta2} \end{align} Finally \begin{align} \frac{|BD|}{|FD|} &= \frac{8\sin^2\tfrac\beta2}{b} \cdot \frac{a}{2\sin\tfrac\beta2} = \frac{2\sin\tfrac\beta2}{ \tfrac{b}2/a } =2. \end{align}