Question

Let ABC be a triangle with centroid G and incentre I.
Statement-1 : If GI is parallel to the side CA, then a, b, c are in A.P. because
Statement-2 : In a triangle, incentre from the angular point A divides the angle bisector in the ratio of $a:(b+c)$ reckoning from the vertex.

• A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
• B. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
• C. Statement-1 is true, statement-2 is false.
• D. Statement-1 is false, statement-2 is true

Answer 1

Answer: (C)

Let the median and the internal bisector of angle through B meet the side CA at D and E respectively. GI is parallel to CA
Here $△BGI\sim △BDE.$ (corresponding sides of similiar triangles arbproportional)

$\therefore \frac{BG}{GD}=\frac{2}{1}=\frac{BI}{IE}=\frac{AB}{AE}=\frac{a+c}{b}$
$\therefore \frac{2}{1}=\frac{c+a}{b}$
$\Rightarrow c+a=2b,a,b,c\text{ are in A.P. }$
$S-2$ is false and it should be $\frac{b+c}{a}$