Question

$ABCD$ is a parallelogram.

Show that $2DC-DB=AC$.

Answer 1

Show that $2DC-DB=AC$.

In the parallelogram given above according to the property of the parallelogram that is the opposite sides of a parallelogram are equal. we can say that,
$$\begin{gathered} AB=DC(\mathrm{opposite}\mathrm{sides}\mathrm{of}\mathrm{parallelogram}) \\ AD=BC(\mathrm{opposite}\mathrm{sides}\mathrm{of}\mathrm{parallelogram}) \end{gathered}$$

Now, we will use the triangle law of vector addition which says that when two vectors are represented as the two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and the direction of the resultant vector.
So, if we use the triangle law of vector addition here, we can clearly say that,
$DA+AB=DB...\left(1\right)$
And also,
$-DA+DC=AC$ (here we have used $-DA$ instead of $DA$ because of the opposite direction of vector that is negative direction of the vector)
So, we can say that,

$DC=AC+DA...\left(2\right)$

Now, we are given that,

$$\begin{gathered} 2DC–DB=2(\mathrm{AC}+\mathrm{DA})–(\mathrm{DA}+\mathrm{AB})\left[\mathrm{from}\left(1\right)\mathrm{and}\left(2\right)\right] \\ =2\mathrm{AC}+2\mathrm{DA}–\mathrm{DA}–\mathrm{AB} \\ =2\mathrm{AC}+\mathrm{DA}–\mathrm{AB} \\ =2\mathrm{AC}+\mathrm{DA}–\mathrm{CD}\left[\because \mathrm{AB}=\mathrm{CD}\right] \\ =2\mathrm{AC}+(-\mathrm{AC})\left[\mathrm{from}\left(2\right)DC=AC+DA\right] \\ =\mathrm{AC} \end{gathered}$$

Hence, it is proved that $2DC-DB=AC$.