Question

(Ceva’s theorem ) ABC is a triangle and AX, BY and CZ are three concurrent cevians. Then prove:
$\dfrac{BX}{XC}\cdot \dfrac{CY}{YA}\cdot \dfrac{AZ}{ZB}=1$

Answer 1

Hint:
Here it is given that AX, BY and CZ are three concurrent cevians of a triangle ABC. Then point X, Y, Z are the midpoint of the sides BC, AC and AB respectively. Therefore, the ratio of BX and XC, CY and YA, AZ and ZB will be equal to one. Thus, the product of all these ratios will be equal to one.

Complete step by step solution:
It is given that AX, BY and CZ are three concurrent cevians of a triangle ABC. Then, the point X is a midpoint of side BC, the point Y is a midpoint of side AC, and the point Y is a midpoint of the side AB.
Thus, length of BX and XC will be half of the side BC, length of AY and YC will be half of the side BC and length of BZ and AZ will be half of the side BC.
We will write it mathematically now.
$\begin{align}
  & BX=XC=\dfrac{1}{2}BC...........\left( 1 \right) \\
 & YA=CY=\dfrac{1}{2}AC...............\left( 2 \right) \\
 & AZ=ZB=\dfrac{1}{2}AB..................\left( 3 \right) \\
\end{align}$
Now, we will find the ratio of$\dfrac{BX}{XC}$ ,$\dfrac{CY}{YA}$and$\dfrac{AZ}{ZB}$.
We will first the ratio of $\dfrac{BX}{XC}$by putting the value of BX and XC from equation 1.
$\dfrac{BX}{XC}=\dfrac{\dfrac{1}{2}BC}{\dfrac{1}{2}BC}=1......\left( 4 \right)$
We will first the ratio of$\dfrac{CY}{YA}$ by putting the value of CY and YA from equation 2.
$\dfrac{CY}{YA}=\dfrac{\dfrac{1}{2}AC}{\dfrac{1}{2}BC}=1..............\left( 5 \right)$
We will first the ratio of$\dfrac{AZ}{ZB}$ by putting the value of AZ and ZB from equation 3.
$\dfrac{AZ}{ZB}=\dfrac{\dfrac{1}{2}AB}{\dfrac{1}{2}AB}=1.............\left( 6 \right)$
We will multiply equation 1, equation 2, and equation 3 now.
$\begin{align}
  & \dfrac{BX}{XC}\cdot \dfrac{CY}{YA}\cdot \dfrac{AZ}{ZB}=1\times 1\times 1=1 \\
 & \therefore \dfrac{BX}{XC}\cdot \dfrac{CY}{YA}\cdot \dfrac{AZ}{ZB}=1 \\
\end{align}$

Hence, we have proved the ceva’s theorem.

Note:
This theorem is known as Ceva’s theorem. Let’s consider the same triangle to define the ceva’s theorem. According to this theorem, if AX, BY and CZ are three concurrent cevians then $\dfrac{BX}{XC}\cdot \dfrac{CY}{YA}\cdot \dfrac{AZ}{ZB}=1$. The converse of this theorem is also true i.e. if$\dfrac{BX}{XC}\cdot \dfrac{CY}{YA}\cdot \dfrac{AZ}{ZB}=1$, then AX, BY and CZ will be three concurrent cevians.