1. Tardigrade
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  4. 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.

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Q. 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.

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Solution:

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 $\triangle B G I \sim \triangle B D E . $ (corresponding sides of similiar triangles arbproportional)
image
$\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 } $
$\Rightarrow c + a =2 b , a , b , c \text { are in A.P. }$
$S -2$ is false and it should be $\frac{ b + c }{ a }$