## Concurrent Segments In A Triangle

#### Problem

Consider triangle $ABC$ and the segments joining the points $X$, $Y$, and $Z$, on opposite edges.

Prove that the segments $AX$, $BY$, and $CZ$ are concurrent at $P$ if and only if $\dfrac{AZ}{BZ}\dfrac{BX}{CX}\dfrac{CY}{AY} = 1$.

#### Solution

This result is known as Ceva's Theorem and is a fundamental theorem used in advanced Euclidean geometry. We shall prove it in two different ways. The first method uses area and the second method makes use of adding an auxillary line.

**Area Method**

Considering $AC$ as the base of $\Delta ABC$, let the altitude be $h$.

$\therefore A\Delta ABY = \dfrac{AY \times h}{2} \implies AY = \dfrac{2 \times A \Delta ABY}{h}$; similarly $CY = \dfrac{2 \times A\Delta CBY}{h}$.

Hence $\dfrac{AY}{CY} = \dfrac{\dfrac{2 \times A\Delta ABY}{h}}{\dfrac{2 \times A \Delta CBY}{h}} = \dfrac{A \Delta ABY}{A \Delta CBY}$.

In the same way, $\dfrac{AY}{CY} = \dfrac{A \Delta APY}{A \Delta CPY}$.

Using the fact that $\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{a - c}{b - d}$ (see Proportional Difference)

$\dfrac{AY}{CY} = \dfrac{A \Delta ABY - A \Delta APY}{A \Delta CBY - A \Delta CPY} = \dfrac{A \Delta ABP}{A \Delta CBP}$

By considering edges $AB$ and $BC$ in the same way we get:

$\dfrac{AZ}{BZ} = \dfrac{A \Delta ACP}{A \Delta BCP}$ and $\dfrac{BX}{CX} = \dfrac{A \Delta BAP}{A \Delta CAP}$

$\therefore \dfrac{AZ}{BZ}\dfrac{BX}{CX}\dfrac{CY}{AY} = \dfrac{A \Delta ACP}{A \Delta BCP}\dfrac{A \Delta BAP}{A \Delta CAP}\dfrac{A \Delta CBP}{A \Delta ABP} = 1$

Hence we have proved that if $AX$, $BY$, and $CZ$ are concurrent then the product of given ratios will be one.

**Auxillary Line Method**

Consider the following diagram where a line parallel to $AC$ has been added and $AX$ has been extended to meet this line at $M$ and $CZ$ has been extended to $N$.

As $\angle BXM = \angle AXC$ (opposite angles), $\angle CAM = \angle BMX$ (alternate angles), and $\angle ACX = \angle MBX$ (alternate angles), $\Delta AXC \sim \Delta BXM$ (similar triangles). By considering this pair of similar triangles and three other pairs of similar triangles we can derive four ratios.

$\Delta AXC \sim \Delta BXM \implies \dfrac{BX}{CX} = \dfrac{BM}{AC}$

$\Delta AZC \sim \Delta BZN \implies \dfrac{AZ}{BZ} = \dfrac{AC}{BN}$

$\Delta CPY \sim \Delta BPN \implies \dfrac{CY}{BN} = \dfrac{PY}{BP}$

$\Delta APY \sim \Delta BPM \implies \dfrac{AY}{BM} = \dfrac{PY}{BP}$

From the last two ratios we get $\dfrac{CY}{NB} = \dfrac{AY}{BM} \implies \dfrac{CY}{AY} = \dfrac{BN}{BM}$.

Multiplying the first two ratios by this new ratio we get the desired result:

$\dfrac{BX}{CX}\dfrac{AZ}{BZ}\dfrac{CY}{AY} = \dfrac{BM}{AC}\dfrac{AC}{BN}\dfrac{BN}{BM} = 1$

Now it is necessary to prove the converse: if the product of ratios are equal to one then the segments are concurrent.

Suppose that $AX$ and $CZ$ intersect at a common point $P$. Construct $BY'$ that passes through $P$.

As these segments are concurrent by construction it follows that $\dfrac{AZ}{BZ}\dfrac{BX}{CX}\dfrac{CY'}{AY'} = 1$.

However, if we move $Y$ along $AC$ such that $\dfrac{AZ}{BZ}\dfrac{BX}{CX}\dfrac{CY}{AY} = 1$, then it follows that $\dfrac{CY}{AY} = \dfrac{CY'}{AY'}$ and $Y$ must coincide with $Y'$ proving concurrency.