Consider a triangle $ABC,$ and let $D$ the foot of the bisector of angle $\angle{BAC},$ $M$ the midpoint of $BC,$ $D'$ the symmetric of $D$ with respect to $M.$ Consider the circle $\omega_1$ tangent externally to $BC$ in $D$ and internally to $\Gamma,$ the circumcircle of $ABC.$ Similarly consider the circle $\omega_2$ tangent externally to $BC$ in $D'$ and internally to $\Gamma.$ Let $E$ be the center of $\omega_1$ and let $E'$ be the center of $\omega_2.$ Prove that the following relation holds:
$$(\overline{AB}-\overline{AC})^2=\overline{AE'}^2-\overline{AE}^2$$
All my attempts didn't give anything. I tried using coordinates but without any result; without coordinates, I know that the line between the center $O$ of $\Gamma$ and the center of $\omega_1$ intersect $\Gamma$ in the tangency point, and similarly for $\omega_2…$
Can anyone give me at least one idea on how to prove this fact?
Best Answer
First of all, we note that the two circles are identical/symmetrical with respect to the perpendicular bisector of $BC$. So, the radius of $\omega_1$ is equal to that of $\omega_2$, and $BE=CE'$. In addition, $\angle EBD=\angle E'CD'=\theta.$ And, let's assume $AB=c, BC=a$, and $AC=b.$
Now, we have:
$$AE^2=c^2+BE^2-2BE.c.\cos (B+\theta); \\ AE'^2=b^2+CE'^2-2CE'.b.\cos (C+\theta).$$
Thus, $$AE'^2-AE^2=b^2-c^2-2BE.b.\cos C\cos \theta+2.BE.b.\sin C\sin\theta\\ +2BE.c.\cos B\cos \theta-2BE.c.\sin B\sin \theta;$$
since $b\sin C=c\sin B$, and $\cos \theta =\frac{BD}{BE}$, we conclude that:
$$AE'^2-AE^2=b^2-c^2-2BD.b.\cos C+2BD.c.\cos B. $$
On the other hand, $BD=\frac{ca}{c+b}$. So, we must show that:
$$b^2-c^2-2BD.b.\cos C+2BD.c.\cos B\\ =b^2-c^2-2\frac{abc}{c+b}\cos C+2\frac{c^2a}{c+b}\cos B=(AB-AC)^2=(c-b)^2,$$
or equivalently,
$$c^2=bc+\frac{c^2a}{c+b}\cos B-\frac{abc}{c+b}\cos C \\ \iff c^2-b^2=ca\cos B-ab\cos C \\ \iff a^2-2ab\cos C=ca\cos B-ab\cos C \\ \iff a=c\cos B+b\cos C,$$
which is a well-known relation in a triangle.
Hence, we are done.