Congruent Segments Associated to Tangent Lines.

Let $A$ and $B$ be the centers two circles $(A)$, $(B)$. From $A$, draw a tangent to $(B)$ at $C$. From $B$, draw two tangents to $(A)$ in $D$, $E$. From $C$, draw $CD$ such that intersect $(A)$ again in $F$. Similarly, from $C$, draw $CE$ such that intersect $(A)$ in $I$.

Prove $CF = CE$.

Proof.

Since $BD$ and $AC$ are tangent lines it follows that $\angle{BDA}=\angle{ACB}=\angle{AEB}=90^\circ$, hence $A$, $B$, $C$, $D$, $E$ are concyclic. Now, see that $\angle{DCE}=\angle{DBE}$ and $\angle{ACE}=\angle{ABE}$. $\angle{ABE}=\frac{\angle{DBE}}{2}=\frac{\angle{DCE}}{2}$, then, segment $AC$ is an angle bisector of $\angle{DCE}$. Focus on $\triangle{ACF}$. $\angle{CDB}=\angle{CEB}$. $\angle{FDA}=\angle{DFA}=180^\circ-(90^\circ+\angle{CDB})=90^\circ-\angle{CDB}$.$\angle{FAC}= 180^\circ-\angle{DCA}-(90^\circ-\angle{CDB})=90^\circ+\angle{CDB}-\angle{DCA}$. Focus on $\triangle{ACE}$. $\angle{EAC}=180^\circ-\angle{DCA}-(90^\circ-\angle{CDB})$. $\angle{EAC}=90^\circ+\angle{CDB}-\angle{DCA}$. Hence $\angle{FAC}=\angle{EAC}$, and $\triangle{ACF}$, $\triangle{ACE}$ are congruent. Thus, $CF=CE$.

On a constant associated to equilateral triangle and its generalization

I guess you are familiar with the result described as follows:

If $ABC$ is an equilateral triangle, and $P$ is any point on the incircle of $\triangle{ABC}$, then $AP^2 + B^2 + CP^2$ is constant. Click here to read more.

I was wondering whether this can be true for any regular polygon and found it to be true too. In the website it is mentioned that the result holds for any circle with center at the centroid of the triangle, but it does not mention whether the result holds for any other polygon. Although it seems a natural question, I have not seen any reference so far.

Click here for the generalization.

Relationship Between Inradius and Exradii

Prove $\frac{r}{r_a}+\frac{r}{r_b}+\frac{r}{r_c} = 1$

I am grateful to Angel Rafael Mejia for the reference:

Johnson, R. A. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, 1929.

The Three Square Geometry Problem

This is my solution of the famous Three Square Geometry Problem. 

Click on the image to have a better view.

A Perspector Associated to Cevians

This is an old open problem by  Oai Thanh Đào. Here I give a proof.

Consider a triangle ABC. Let D be a point inside triangle ABC. Let EFG be the cevian triangle of point D. On segments BF, CF, CG, AG, AE, BE erect similar isosceles triangles BFH, CFI, CGJ, AGK, AEL and BEM, respectively. Let N be the intersection of HI, LM. Similarly, let O be the intersection of HI, KJ and P the intersection of KJ, ML.  Then, AP, BN, CO concur. 

Proof (click on the image to have a better view).

Related topics:

A family of perspectors associated to cevians

Pythagoras in 60 Seconds

Short film by Alan Kitching, made in 1986. Apart from explaining the famous theorem, it is also conceived as a "visual proof"... ie, whereas a still diagram can show a single instance of something, animation can show all possible instances... and that can make the difference between a simple illustration and an actual proof.

Another Archimedean circle in an arbelos

Click here to read the article.