Relation between tangent circles and side of a triangle

Question

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: $$(\overline{AB}-\overline{AC})^2=\overline{AE'}^2-\overline{AE}^2$$

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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?

Answer 1

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.

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Answer 2

Here is an approach in two steps.

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Claim: $\overline{AE'}^2-\overline{AE}^2=\overline{AD'}^2-\overline{AD}^2$

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Proof. The two smaller circles are symmetrical to the line through $M$ that is perpendicular to $DD'$. Let us set up the coordinate system so that we have $E(-a,0)$, $E'(a,0)$, $D(-a, -r)$, $D'(a, -r)$ and $A(p,q)$. Then $$\begin{aligned} \overline{AE'}^2-\overline{AE}^2&=((p-a)^2-(q-0)^2)-((p+a)^2-(q-0)^2)=(p-a)^2-(p+a)^2\\ \overline{AD'}^2-\overline{AD}^2&=((p-a)^2-(q+r)^2)-((p+a)^2-(q+r)^2)=(p-a)^2-(p+a)^2\\ \end{aligned}$$

(No special property of $A$ is used for this claim.)

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Claim: $\overline{AD'}^2-\overline{AD}^2=(\overline{AB}-\overline{AC})^2$

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Proof. Let $\overline {BA}=c$, $\ \overline{CB}=a$, $\ \overline{AC} =b$. Then $\overline{BD}=\frac c{b+c}a$, $\ \overline{BD'}=\overline{CD}=\frac{b}{b+c}a$.

$$\begin{aligned} &\quad\overline{AD'}^2-\overline{AD}^2\\ &=(\overline{BA}^2 + \overline{BD'}^2-2\overline{BA}\,\overline{BD'}\cos B)-(\overline{BA}^2 + \overline{BD}^2-2\overline{BA}\,\overline{BD}\cos B)\\ &=(\overline{BD'}-\overline{BD})(\overline{BD'}+\overline{BD}-2\overline{BA}\cos B)\\ &=\frac{b-c}{b+c}a(a-2c\cos B)\\ &=\frac{b-c}{b+c}(a^2-2ac\frac{a^2+c^2-b^2}{2ac})\\ &=(b-c)^2\\ &=(\overline{AB}-\overline{AC})^2 \end{aligned}$$

Answer 3

Fig. 1: $\textit{Don't you find a resemblance with Humpty Dumpty ?}$

First of all, the two little circles are clearly symmetrical with respect to the line bisector $MF$ of line segment $BC$. In particular $ED$ and $E'D'$ are symmetrical with respect to this line.

With usual notations $a=BC,b=AC, c=AB$, the relationship to be established is :

$$(c-b)^2=\overline{AE'}^2-\overline{AE}^2.\tag{1}$$

Let us work on the RHS of (1) written in an equivalent way with vectors :

$$\text{rhs}=\vec{AE'}^2-\vec{AE}^2\tag{2}$$

$$\text{rhs}=(\vec{AE'}-\vec{AE}).(\vec{AE'}+\vec{AE})\tag{3}$$

(where the point means "dot product").

$$\text{rhs}=\vec{EE'}.(\vec{AD'}+\vec{D'E'}+\vec{AD}+\vec{DE})\tag{4}$$

As $\vec{EE'} \perp \vec{D'E'}$ and $\vec{EE'} \perp \vec{DE}$, (4) can be reduced to:

$$\text{rhs}=\vec{EE'}.(\vec{AD'}+\vec{AD})\tag{5}$$

$$\text{rhs}=\vec{DD'} . (2 \vec{AM})\tag{6}$$

Let $H$ be the projection of $A$ onto line $BC$ ; (6) can be written :

$$\text{rhs}=2 \vec{DM} . 2 (\vec{AH}+ \vec{HM})\tag{7}$$

By orthogonality of $\vec{DM}$ and $\vec{AH}$, (7) boils down to

$$\text{rhs}=4\vec{DM}.\vec{HM}\tag{8}$$

But a separate result on which I have worked and that I have given as a self-contained question/answer here establishes that this RHS of (1) is equal to the corresponding LHS, i.e., $(b-c)^2.$