Show that the area of a parallelogram whose two adjacent edges are two diagonals of a given parallelogram is double the area of given parallelogram.

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Solution

Let $A B C D$ be a parallelogram.
Let $\to a$ and $\to b$ represents the adjacent edges of parallelogram ABCD.
So, $\to A C = \to a$ and $\to B D = \to b$
Let the diagonals intersect at point E.
Since, the diagonals of a parallelogram bisect each other at E.
$\to A E = \to E C = \frac{\to a}{2}$
$\to B E = \to E D = \frac{\to b}{2}$
Now , in triangle $A E B$ ,
$\to A B = \to A E + \to E B$
$= \frac{\to a}{2} - \frac{\to b}{2}$
$\Rightarrow \to A B = \frac{1}{2} (\to a - \to b)$

Now , in triangle $A E D$ ,
$\to A D = \to A E + \to E D$
$= \frac{\to a}{2} + \frac{\to b}{2}$
$\Rightarrow \to A B = \frac{1}{2} (\to a + \to b)$

We know that
Area of parallelogram $= | A B \times A D |$
$= | \frac{1}{2} (\to a - \to b) \times \frac{1}{2} (\to a + \to b) |$
$= | \frac{1}{4} (\to a - \to b) \times (\to a + \to b) |$
$= \frac{1}{4} | \to a \times \to a + \to a \times \to b - \to b \times \to a + \to b \times \to b |$
$= \frac{1}{4} | 0 + \to a \times \to b + \to a \times \to b + 0 |$
$= \frac{1}{4} | 2 (\to a \times \to b) |$
$= \frac{1}{2} | \to a \times \to b |$

So, the area of parallelogram with diagonals $\to a$ and $\to b$ is $\frac{1}{2} | \to a \times \to b |$ , which is half the area of given parallelogram

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