Question

In the adjoining figure, the medians BD and CE of a $\Delta ABC$ meet at $G$ then $BG=2\times GD$

• A. True
• B. False

Answer 1

The correct option is A True

Since, $BD$ and $CE$ are medians

$AD=DC$

$AE=BE$

Hence, by converse of Basic Proportionality theorem,

$ED\parallel BC$

In $△EGD$ and $△CGB,$

$\Rightarrow$ $\angle DEG=\angle GCB$ [ Alternate angles ]

$\Rightarrow$ $\angle EGD=\angle BGC$ [ Vertically opposite angles ]

$\Rightarrow$ $△EGD\sim △CGB$ [ $AA$ similarity test ]

$\therefore$ $\frac{GD}{GB}=\frac{ED}{BC}$ ----- ( 1 )

In $△AED$ and $△ABC,$

$\Rightarrow$ $\angle AED=\angle ABC$ [ Corresponding angles ]

$\Rightarrow$ $\angle EAD=\angle BAC$ [ Common angle ]

$\Rightarrow$ $△EAD\sim △BAC$ [ $AA$ similarity test ]

$\therefore$ $\frac{ED}{BC}=\frac{AE}{AB}=\frac{1}{2}$ [ Since, $E$ is the mid-point of $AB$ ]

$\Rightarrow$ $\frac{ED}{BC}=\frac{1}{2}$

From ( 1 ),

$\Rightarrow$ $\frac{GD}{GB}=\frac{1}{2}$

$\Rightarrow$ $GB=2GD$