In an isoceles triangle ABC with …

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