Question
Prove that medians of a triangle are concurrent
✓
Answer
Consider $\triangle ABC$.
Let $P, Q, R$ be the midpoints of the sides $B C, C A, A B$ respectively.
Let the medians $B Q$ and $C R$ intersect at $G$.
To prove that the third median AP also passes through $G$.
Let $\overline{ a }, \overline{ b }, \overline{ c }, \overline{ p }, \overline{ q }, \overline{ r }, \overline{ g }$ be the position vectors of the points $A, B, C, P, Q , R , G$ respectively.
Since $P, Q, R$ are the mid-points of the sides $B C, C A, A B$ respectively
$\therefore$ By midpoint formula, we get
$\overline{ p }=\frac{\overline{ b }+\overline{ c }}{2}\ldots(i)$
$\overline{ q }=\frac{\overline{ c }+\overline{ a }}{2}\ldots(ii)$
$\overline{ r }=\frac{\overline{ a }+\overline{ b }}{2}\ldots(iii)$
From (i), (ii) and (iii), we get
$2 \overline{ p }=\overline{ b }+\overline{ c } \Rightarrow 2 \overline{ p }+\overline{ a }=\overline{ a }+\overline{ b }+\overline{ c }$
$2 \overline{ q }=\overline{ c }+\overline{ a } \Rightarrow 2 \overline{ q }+\overline{ b }=\overline{ a }+\overline{ b }+\overline{ c }$
$2 \overline{ r }=\overline{ a }+\overline{ b } \Rightarrow 2 \overline{ r }+\overline{ c }=\overline{ a }+\overline{ b }+\overline{ c }$
$\therefore \frac{2 p +\overline{ a }}{3}=\frac{2 \overline{ q }+\overline{ b }}{3}=\frac{2 \overline{ r }+\overline{ c }}{3}=\frac{\overline{ a }+\overline{ b }+\overline{ c }}{3}$
$\therefore \frac{2 p +\overline{ a }}{2+1}=\frac{2 \overline{ q }+\overline{ b }}{2+1}=\frac{2 \overline{ r }+\overline{ c }}{2+1}=\frac{\overline{ a }+\overline{ b }+\overline{ c }}{3}$
$=\overline{ g } \quad \ldots \ldots . \text { (say) }$
This shows that the point $G$ whose position vector is $\bar{g}$ lies on the three medians $A P, B Q, C R$ dividing them internally in the ratio $2: 1$.
Hence, the three medians are concurrent.
Let $P, Q, R$ be the midpoints of the sides $B C, C A, A B$ respectively.
Let the medians $B Q$ and $C R$ intersect at $G$.
To prove that the third median AP also passes through $G$.
Let $\overline{ a }, \overline{ b }, \overline{ c }, \overline{ p }, \overline{ q }, \overline{ r }, \overline{ g }$ be the position vectors of the points $A, B, C, P, Q , R , G$ respectively.
Since $P, Q, R$ are the mid-points of the sides $B C, C A, A B$ respectively
$\therefore$ By midpoint formula, we get
$\overline{ p }=\frac{\overline{ b }+\overline{ c }}{2}\ldots(i)$
$\overline{ q }=\frac{\overline{ c }+\overline{ a }}{2}\ldots(ii)$
$\overline{ r }=\frac{\overline{ a }+\overline{ b }}{2}\ldots(iii)$
From (i), (ii) and (iii), we get
$2 \overline{ p }=\overline{ b }+\overline{ c } \Rightarrow 2 \overline{ p }+\overline{ a }=\overline{ a }+\overline{ b }+\overline{ c }$
$2 \overline{ q }=\overline{ c }+\overline{ a } \Rightarrow 2 \overline{ q }+\overline{ b }=\overline{ a }+\overline{ b }+\overline{ c }$
$2 \overline{ r }=\overline{ a }+\overline{ b } \Rightarrow 2 \overline{ r }+\overline{ c }=\overline{ a }+\overline{ b }+\overline{ c }$
$\therefore \frac{2 p +\overline{ a }}{3}=\frac{2 \overline{ q }+\overline{ b }}{3}=\frac{2 \overline{ r }+\overline{ c }}{3}=\frac{\overline{ a }+\overline{ b }+\overline{ c }}{3}$
$\therefore \frac{2 p +\overline{ a }}{2+1}=\frac{2 \overline{ q }+\overline{ b }}{2+1}=\frac{2 \overline{ r }+\overline{ c }}{2+1}=\frac{\overline{ a }+\overline{ b }+\overline{ c }}{3}$
$=\overline{ g } \quad \ldots \ldots . \text { (say) }$
This shows that the point $G$ whose position vector is $\bar{g}$ lies on the three medians $A P, B Q, C R$ dividing them internally in the ratio $2: 1$.
Hence, the three medians are concurrent.
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