Question 435666
Let A , B, and C be the interior angles of the triangle, and a, b, and c, their respective opposite sides.

Then the area of the triangle ABC is given by 

{{{A = (1/2)ab*sinC}}}.

Now from the sine law, {{{a/sinA = c/sinC}}}==> {{{a = c(sinA/sinC)}}}, and 
{{{b/sinB = c/sinC}}}==> {{{b = c(sinB/sinC)}}}.  Then 

{{{A =(1/2)ab*sinC = (1/2)*c*(sinA/sinC)*c(sinB/sinC)*sinC
= (c^2/2)((sinA*sinB)/sinC)

}}}.
But C = 180 - (A+B)==> sinC = sin(180-(A+B)) = sin(A+B), whence
{{{A = (c^2/2)((sinA*sinB)/sinC) = (c^2/2)((sinA*sinB)/sin(A+B)), 
and the proof is complete.

}}}.