PRMO 2012 question 8

 

In rectangle \(ABCD\), \(AB = 5\) and \(BC = 3\). Points \(F\) and \(G\) are on line segment \(CD\) so that \(DF = 1\) and \(GC = 2\). Lines \(AF\) and \(BG\) intersect at \(E\). What is the area of \(\triangle AEB\)?

Solution:
Triangles EAB and EFG are similar by AAA.
Let's say height of EFG is h.
Then height of EAB is 3 + h.
So AB/FG = (3+h)/h
=> 5/2 = (3+h)/h
=> 5h = 6 + 2h
=> h = 2
=> 3 + h = 5
So area of EAB = (5*5)/2 = 12.5



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