Let AC and CE be perpendicular line segments,each of length 18.Suppose B and D are the mid points of AC and CE respectively. If F is the intersection of EB and AD ,then the area of triangle DEF is
My attempt is Area of $\Delta \space DEF=($ Area of $\Delta \space ACD$+ Area of $\Delta \space BCE$ – Area of quadrilateral BCDF)/2
as Triangles ABF and DEF are congruent.
Now Quadrilateral BCDEF can be divided by line segment CF into $\Delta \space BCF \space and \space \Delta CFD$
Now $\angle ADC = tan^{-1}(9/18) \implies tan^{-1}(\frac{1}{2}) $
Similarly we can also get $\angle CBF$ and as $\Delta BCF$ is congruent to $\Delta DCF$ , $\angle BFC \space = \angle DFC$ and as $\angle BCF \space = \angle DCF= 45^{\circ}$ , we can get $\angle BFC \space and \space \angle DFC$
Finally using the sine rule we can get BF,CF,FD,CD using which we can calculate the area of the quadrilateral BCDF as a sum of triangles BFC and DFC
We can easily get the area of triangles ACD and BCE as they are right angled triangles and subtracting it by the new found area of quadrilateral BCDF, we can get our result.
I do not want this process. I want an easier approach to this problem. Please suggest with reasons.
Best Answer
Join the points $C$ and $F$
The area of triangle $CFD$ is the same as the area of triangle $FED$ since they have the same base and height. Also by symmetry, area $BFC$ is equal to each of these.
Can you finish?