Question

The medians of a triangle $ABC$ intersect each other at point $G$ . If one of its medians is $AD$ .prove that Area $(△BGC)=\frac{1}{3}\times$ Area $(△ABC)$

Answer 1

Given, the medians of a triangle ABC intersect each other at point G . One of its medians is AD
Using property when the medians of triangles intersect each other, it divides into ratio of 2:1.
So, AG:GD$=$ 2:1
or, GD$=\frac{1}{3}$ AD (1)
Drawing a altitude BF from B to AD at F.
So,Area of $△ABD$$=\frac{1}{2}\times base\times altitude=\frac{1}{2}\times AD\times BF$
Area of $△BGD$$=\frac{1}{2}\times base\times altitude=\frac{1}{2}\times GD\times BF$
$=\frac{1}{2}\times \frac{1}{3}AD\times BF$ ( GD$=\frac{1}{3}$ AD, From 1)
Taking the ratio of both the areas ,
$\frac{Areaof△ABD}{Areaof△BGD}=\frac{\frac{1}{2}\times AD\times BF}{\frac{1}{2}\times \frac{1}{3}AD\times BF}=3$
or, Area of $△ABD$$=3\times$ Area of $△BGD$
or,Area of $△BGD$$=\frac{1}{3}\times$ Area of $△ABD$ (2)
Again,Drawing a altitude CF from C to AD at F.
So,Area of $△ACD$$=\frac{1}{2}\times base\times altitude=\frac{1}{2}\times AD\times CF$
Area of $△CGD$$=\frac{1}{2}\times base\times altitude=\frac{1}{2}\times GD\times CF$
$=\frac{1}{2}\times \frac{1}{3}AD\times CF$ ( GD$=\frac{1}{3}$ AD, From 1)
Taking the ratio of both the areas ,
$\frac{Areaof△ACD}{Areaof△CGD}=\frac{\frac{1}{2}\times AD\times CF}{\frac{1}{2}\times \frac{1}{3}AD\times CF}=3$
or, Area of $△ACD$$=3\times$ Area of $△CGD$
or, Area of $△CGD$$=\frac{1}{3}\times$ Area of $△ACD$ (3)
Adding equation (2) and (3),
Area of $△BGD$ + Area of $△CGD$$=$$\frac{1}{3}\times$ Area of $△ABD$ + $\frac{1}{3}\times$ Area of $△ACD$
So, Area of $△BGC$$=\frac{1}{3}\times$(Area of $△ABD$ + Area of $△ACD$)
or, Area of $△BGC$$=$$\frac{1}{3}\times$ Area of $△ABC$