# Relation between tangent circles and side of a triangle

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?

• Have you noticed that $\omega$ and $\omega'$ are symmetrical with respect to the perpendicular bissector of $BC$ ? Mar 29 at 21:34
• Yes, I've noticed this fact, but I don't know how to use it... Maybe could It be useful to reflect also the point A? My first problem is on how to "put together" the segments in the thesis, to find some relations between them... Mar 29 at 21:52
• I have some ideas (without having completely "bridged the gap" between LHS and RHS), but can you say, as you have tagged "contest-math", which kind of contest/olympiad is this problem coming from ? Mar 29 at 23:49

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. • Interesting proof. Please have a look at mine, that I have just posted : it uses vector computations instead of trigonometry. Mar 31 at 12:50 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., $$(b-c)^2.$$

• Nice observation with a relatively small amount of calculation! [+1] Mar 31 at 17:40

Here is an approach in two steps.

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

$$\qquad$$ 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.)

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

$$\qquad$$ 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}

• Please have a look at the proof I just posted (there are common points with your interesting proof, but expressed differently). Mar 31 at 12:48