Answer:
The area of △AON is one sixth part of ABC.
Step-by-step explanation:
Let the area of triangle ABC be x.
It is given that AM and CN are median and they are intersecting at point O. So point and is centroid of the triangle.
We know that centroid divides the medians in 2:1. Therefore the CO:ON is 2:1.
Since CN is median, therefore it divides the area of triangle in two equal parts.
[tex]\text{Area of }\triangle ACN=\text{Area of }\triangle BCN=\frac{x}{2}[/tex]
Since the point O divides the line CN is 2:1, therefore the line AO divides the area of triangle ACN is 2:1.
Draw a perpendicular from A to CN. The length of AH is h.
[tex]\frac{\text{Area of }\triangle AON}{\text{Area of }\triangle ACN}=\frac{\frac{1}{2}\times ON\times h}{\frac{1}{2}\times CN\times h}=\frac{1}{3}[/tex]
Therefore the area of triangle AON is one-third of area of triangle ANC.
[tex]\text{Area of }\triangle AON=\frac{1}{3}\times \frac{x}{2}[/tex]
[tex]\text{Area of }\triangle AON=\frac{x}{6}[/tex]
[tex]\frac{\text{Area of }\triangle AON}{\text{Area of }\triangle ABC}=\frac{(\frac{x}{6})}{x}=\frac{1}{6}[/tex]
Therefore, the area of △AON is one sixth part of ABC.
