Find the ratio of the area of triangle BCX to the area of triangle ACX in the diagram if CX bisects the angle ACB. Express your answer as a common fraction. This is the image https://latex.artofproblemsolving.com/2/4/1/2415560a3ecd06119f3e9d785d5d655c158f158e.png
By Euclid, whenever an apex angle is bisected, we have the following relationship :
AX/AC = BX/BC
AX/30 = BX/27 which implies that
27/30 = BX/AX
9 /10 = BX/AX → BX = (9/10) AX
And triangle ACX has the same height as triangle BCX
And again, by Euclid, triangles under the same height are to each other as their bases
So....the area of triangle BCX = (9/10)ACX
And the ratio of their areas is :
BCX : ACX = 9 : 10
Find the ratio of the area of triangle BCX to the area of triangle ACX in the diagram if CX bisects the angle ACB.
Express your answer as a common fraction.
Let \(A_1=\) area of BCX
Let \(A_2=\) area of ACX
Let \(\varphi = \angle ACB\)
Let ratio = \(\frac{A_1}{A_2}\)
\(\begin{array}{|lrcll|} \hline (1) & 2\cdot A_1 &=& \overline{AX}\cdot 27 \cdot \sin(\frac{\varphi}{2}) \\ (2) & 2\cdot A_2 &=& \overline{AX}\cdot 30 \cdot \sin(\frac{\varphi}{2}) \\ \hline \frac{(1)}{(2)}: & \dfrac{2\cdot A_1}{2\cdot A_2} &=& \dfrac{\overline{AX}\cdot 27 \cdot \sin(\frac{\varphi}{2}) }{\overline{AX}\cdot 30 \cdot \sin(\frac{\varphi}{2}) } \\ & \dfrac{ A_1}{ A_2} &=& \dfrac{ 27 }{ 30 } \\ & &=& \dfrac{ 3\cdot 9 }{ 3\cdot 10 } \\ & &=& \dfrac{ 9 }{ 10 } \\\\ &\mathbf{ ratio } & \mathbf{=} & \mathbf{ \dfrac{9}{10} } \\ \hline \end{array}\)