Congruency theorems — lesson. Mathematics State Board, Class 10.

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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 the 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.

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12. Congruency theorems

Theory: