# Prove concurrency (probably using Carnot's Theorem)

Let $ABC$ be a triangle. An arbitrary circle $(K_a)$ passing through $B, C$ intersects $CA, AB$ again at $A_b, A_c$, respectively. Define $B_c, B_a, C_a, C_b$ similarly. Prove that the perpendicular bisector of $C_aB_a, A_bC_b, B_cA_c$ are concurrent.

Hints (from my teacher): Use Carnot Theorem: Let $ABC$ be a triangle . $M,N,P$ are points on $BC, CA, AB$,respectively. Suppose that $\Delta_1,\Delta_2,\Delta_3$ are three lines through $M,N,P$ and perpendicular to $BC,CA,AB$, respectively. Then $\Delta_1;\Delta_2;\Delta_3$ are concurrent iff $MC^2-MB^2+NC^2-NA^2+PA^2-PB^2=0$

Thank you!

• Use the "Power of a Point with respect to a Circle" to relate various segments lengths. For instance, $$|\overline{AA_c}||\overline{AB}| = \text{power of A wrt K_a} = |\overline{AA_b}||\overline{AC}|$$ Also, $|\overline{AA_c}| + |\overline{BA_c}| = |\overline{AB}|$, etc; and $|\overline{AC_c}| = \frac12(|\overline{AB_c}| + |\overline{AA_c}|)$, etc, where $C_c$ is the midpoint of $\overline{A_cB_c}$. – Blue Sep 17 '14 at 21:06
• Can you please explain things more clearly? I'm all messed up because there are too many segment lengths to calculate. – primitiveroot Sep 17 '14 at 22:44

Here's a diagram, where I've marked the feet of the perpendicular bisectors as $A_a$, $B_b$, $C_c$.

As hinted by your teacher, Carnot's Theorem ---not to be confused with Carnot's Theorem :)--- guarantees the desired concurrency once we verify that $$\left(\;|AC_c|^2 - |BC_c|^2\;\right) + \left(\;|BA_a|^2 - |CA_a|^2\;\right) + \left(\;|CB_b|^2 - |AB_b|^2\;\right) \stackrel{?}{=} 0 \tag{\star}$$

Let's consider one part of $(\star)$ at a time: $$|AC_c|^2 - |BC_c|^2 = \left(\;|AC_c|+|BC_c|\;\right)\left(\;|AC_c|-|BC_c|\;\right) = |AB|\;\left(\;|AC_c|-|BC_c|\;\right)$$

Since $$|AC_c| = \frac12 (\; |AA_c| + |AB_c| \;) = \frac12 (\;|AA_c|+|AB|-|BB_c|\;)$$ $$|BC_c| \phantom{= \frac12 (\; |AA_c| + |AB_c| \;)} = \frac12(\;|BB_c|+|AB|-|AA_c|\;)$$ the above becomes $$|AC_c|^2 - |BC_c|^2 = |AB|\;\left(\;|AA_c| - |BB_c|\;\right)$$ Likewise, \begin{align} |BA_a|^2 - |CA_a|^2 &= |BC|\;\left(\;|BB_a|-|CC_a|\;\right) \\ |CB_b|^2 - |AB_b|^2 &= |CA|\;\left(\;|CC_b|-|AA_b|\;\right) \end{align} Therefore, $(\star)$ becomes $$|AB||AA_c| + |BC||BB_a| + |CA||CC_b| \stackrel{?}{=} |BA||BB_c| + |CB||CC_a|+|AC||AA_b| \qquad(\star\star)$$ where I've written "$|AB|$" as "$|BA|$", etc, on the right in anticipation of the following:

The "Power of a Point" theorem says that, if a line through point $P$ meets circle $\kappa$ at points $Q$ and $R$, then $|PQ||QR|$ is a value that depends only on $P$ and $\kappa$, not on $Q$ and $R$. This value is called the power of $P$ with respect to $K$.

In $(\star\star)$, the value of term $|AB||AA_c|$ is the power of point $A$ with respect to $\kappa_A$; but so is the value of $|AC||AA_b|$. These terms cancel. Similarly, $|BC||BB_a| = |BA||BB_c|$ (the power of $B$ with respect to $\kappa_B$), and $|CA|CC_b| = |CB||CC_a|$ (the power of $C$ with respect to $\kappa_C$). All the terms cancel, so that Carnot's Theorem is satisfied: the lines in question are concurrent. $\square$

And here is the elegant proof and the inventor of the problem: http://www.cut-the-knot.org/m/Geometry/GarciaCircles.shtml

• It's always nice to include the essence of the solution here since links may break – Shailesh Jul 30 '16 at 4:09