In an isoceles triangle ABC with …

Given: A quadrilateral ABCD, in which
AC and BD intersect each other at point E.
To Prove: $ar{\text{ }}(AED) \times ar{\text{ }}(BEC)$
$= {\text{ }}ar{\text{ }}(ABE) \times ar{\text{ }}(CDE)$
Construction: A, draw AM $\bot$ BD and from C, draw CN $\bot$ BD.
Proof: ar ($\Delta$ABE) = $\frac{1}{2} \times BE \times AM$ ……….(i)
And ar ($\Delta$AED) = $\frac{1}{2} \times DE \times AM\;$……….(ii)
Dividing eq. (ii) by i), we get,
$\frac{{{\text{ar}}\left( {\Delta {\text{AED}}} \right)}}{{{\text{ar}}\left( {\Delta {\text{ABE}}} \right)}} = \frac{{\frac{1}{2} \times {\text{DE}} \times {\text{AM}}}}{{\frac{1}{2} \times {\text{BE}} \times {\text{AM}}}}$
$\Rightarrow$ ……….(iii)
Similarly  ……….(iv)
From eq. (iii) and (iv), we get
$\frac{{{\text{ar}}\left( {\Delta {\text{AED}}} \right)}}{{{\text{ar}}\left( {\Delta {\text{ABE}}} \right)}}$ = $\frac{{{\text{ar}}\left( {\Delta {\text{CDE}}} \right)}}{{{\text{ar}}\left( {\Delta {\text{BEC}}} \right)}}$
$\Rightarrow ar{\text{ }}(AED) \times ar{\text{ }}(BEC)$
$= {\text{ }}ar{\text{ }}(ABE) \times ar{\text{ }}(CDE)$