### Theory:

Cevian:
A cevian is a line segment that extends from one vertex of a triangle to the opposite side. In the figure, $$AD$$ is a cevian from $$A$$.

Special Cevians of a Triangle:

• A median is a cevian that divides the opposite side into two equal lengths.
• An altitude is a cevian that is perpendicular to the opposite side.
• An angle bisector is a cevian that bisects the corresponding angle.
Ceva’s Theorem (without proof):
Let $$ABC$$ be a triangle and let $$D$$, $$E$$, $$F$$ be points on lines $$BC$$, $$CA$$ and $$AB$$ respectively. Then the cevians $$AD$$, $$BE$$, $$CF$$ are concurrent if and only if $$\frac{BD}{DC} \times \frac{CE}{EA} \times \frac{AF}{FB} = 1$$ where the lengths are directed. This also works for the reciprocal of each of the ratios as the reciprocal of $$1$$ is $$1$$.
Example:
The cevians $$AD$$, $$BE$$ and $$CF$$ of the triangle $$ABC$$ are congruent. If $$\frac{BD}{DC} = \frac{2}{5}$$, $$\frac{CE}{EA} = \frac{3}{4}$$ and $$\frac{AF}{FB} = \frac{x}{6}$$, find the value of $$x$$.

Solution:

By Ceva's theorem, we have $$\frac{BD}{DC} \times \frac{CE}{EA} \times \frac{AF}{FB} = 1$$.

Substitute the required values in the above equation.

$$\Rightarrow$$ $$\frac{2}{5} \times \frac{3}{4} \times \frac{x}{6} = 1$$

$$\Rightarrow$$ $$x$$ $$=$$ $$1 \times \frac{5}{2} \times \frac{4}{3} \times 6$$

$$\Rightarrow$$ $$x$$ $$=$$ $$20$$
Menelaus Theorem (without proof): A necessary and sufficient condition for points $$P$$, $$Q$$, $$R$$ on the respective sides $$BC$$, $$CA$$, $$AB$$ (or their extension) of a triangle $$ABC$$ to be collinear is that $$\frac{BP}{PC} \times \frac{CQ}{QA} \times \frac{AR}{RB} = -1$$ where all segments in the formula are directed segments.
Important!
• Menelaus theorem is also given by $$BP \times CQ \times AR = − PC \times QA \times RB$$.
• If any one of the six directed line segments $$BP$$, $$PC$$, $$CQ$$, $$QA$$, $$AR$$, or $$RB$$ is interchanged, then the product will be $$1$$.
Example:
On a random board, three pins $$P$$, $$Q$$, $$R$$ are punched in the way, $$BP = 3$$ $$cm$$, $$CQ = 4$$ $$cm$$, $$RA = 5$$ $$cm$$, $$PC = 2$$ $$cm$$, $$QA = 5$$ $$cm$$, $$RB = 6$$ $$cm$$, where $$A$$, $$B$$ and $$C$$ are points such that $$P$$ lies on $$BC$$, $$Q$$ lies on $$AC$$ and $$R$$ lies on $$AB$$. Check whether the pins $$P$$, $$Q$$ and $$R$$ lie on a same straight line.

Solution:

Note that the directed segment $$AR$$ is interchanged as $$RA$$.

By Menelaus theorem, the pins $$P$$, $$Q$$ and $$R$$ will lie on the same straight line if BP
$$\frac{BP}{PC} \times \frac{CQ}{QA} \times \frac{AR}{RB} = 1$$.

Consider the left hand side of the equation.

$$\frac{BP}{PC} \times \frac{CQ}{QA} \times \frac{AR}{RB}$$

Substitute the required values in the above equation.

$$=$$ $$\frac{3}{2} \times \frac{4}{5} \times \frac{5}{6}$$

$$=$$ $$\frac{60}{60}$$

$$=$$ $$1$$

Therefore, the pins $$P$$, $$Q$$ and $$R$$ lie on the same straight line.