Question

Statement-1 : Consider triangle ABC and point M is such that triangles ABM, ACM and BCM are of equal area, then point M is centroid of triangle ABC (where A, B, C and M lie in x-y plane)

Statement-2: In triangle ABC, $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = R$, then

(where symbols have their usual meaning)

• A. Statement-1 is true
• B. Statement-2 is true
• C. Statement-1 is false
• D. Statement-2 is false

Answer 1

Answer: (A), (D)

Statement-1 is true, Statement-2 is false

Answer 2

Statement-1 Analysis:

Let the vertices of the triangle be A, B, C, and let M be a point in the x-y plane. We are given that Area(ABM) = Area(ACM) = Area(BCM). Let this common area be $K$. So, Area(ABM) = Area(ACM) = Area(BCM) = $K$.

Consider the line passing through vertex A and point M. Let this line intersect the side BC at a point D'. The ratio of the areas of triangles ABM and ACM can be expressed as the ratio of the segments BD' and CD' on the base BC, because triangles ABM and ACM share the same height from M to the line AD', and triangles ABD' and ACD' share the same height from A to the line BC.

More formally, we know that if a line segment AD' divides the side BC of a triangle ABC, then $\frac{\text{Area(ABD')}}{\text{Area(ACD')}} = \frac{\text{BD'}}{\text{CD'}}$. Similarly, for point M on AD', $\frac{\text{Area(MBD')}}{\text{Area(MCD')}} = \frac{\text{BD'}}{\text{CD'}}$. From these, we can deduce that $\frac{\text{Area(ABM)}}{\text{Area(ACM)}} = \frac{\text{BD'}}{\text{CD'}}$.

Since Area(ABM) = Area(ACM) (given as $K$), we have: $\frac{\text{Area(ABM)}}{\text{Area(ACM)}} = 1$ Therefore, $\frac{\text{BD'}}{\text{CD'}} = 1$, which implies BD' = CD'. This means D' is the midpoint of the side BC. Since the line AM passes through the midpoint of BC, AM is a median of triangle ABC.

Similarly, by considering the line passing through B and M (intersecting AC at E'), we can show that BM is a median of triangle ABC. And by considering the line passing through C and M (intersecting AB at F'), we can show that CM is a median of triangle ABC.

Since point M lies on all three medians of triangle ABC, M must be the point of concurrency of the medians, which is the centroid of the triangle. Hence, Statement-1 is true.

Statement-2 Analysis:

The statement is "In triangle ABC, $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = R$, then (where symbols have their usual meaning)". The phrase "where symbols have their usual meaning" implies that R stands for the circumradius of the triangle. The Sine Rule states that for any triangle ABC,

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$

where 'a', 'b', 'c' are the lengths of the sides opposite to angles A, B, C respectively, and R is the circumradius.

The statement asserts that the common ratio is R, i.e., $\frac{a}{\sin A} = R$. Comparing this with the correct Sine Rule ($\frac{a}{\sin A} = 2R$), we get $R = 2R$. This implies $R = 0$, which is impossible for any triangle (a triangle must have a non-zero circumradius). Therefore, the identity presented in Statement-2 is incorrect if R represents the circumradius. Hence, Statement-2 is false.

Conclusion:

Statement-1 is true. Statement-2 is false.