91Ó°ÊÓ

Assume that diagonals of a nondegenerate quadrangle ABCD intersect at point M. Show that $$\operatorname{area}(\boldsymbol{\Delta} A B M) \cdot \operatorname{area}(\boldsymbol{\Delta} C D M)=\operatorname{area}(\boldsymbol{\Delta} B C M) \cdot \operatorname{area}(\boldsymbol{\Delta} D A M)$$

Let \(r\) be the inradius of \(\triangle A B C\) and \(p\) be its semiperimeter; that is, \(p=\frac{1}{2} \cdot(A B+B C+C A) .\) Show that $$\operatorname{area}(\boldsymbol{\Delta} A B C)=p \cdot r$$