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Please help me solving BDMO 2016: National Junior/ Question No. 2

The Question is:
ABC is a triangular piece of paper with an area of 500 square units and AB = 20 units. DE is parallel to AB. The triangle is folded along DE. The part of the triangle below AB has an area of 80 square units. What is the area of CDE?

Ref.
http://www.matholympiad.org.bd/question ... 6-national

[CDE]=225 sq. units

explain the process please

can u explain please?

Let $CF'G'$ be the reflection of $C'FG$. And as $CDE$ is the reflection of $C'DE$, we can say,
$F'G'||DE||AB$
So, $(CF'G')=(C'FG)=80$.
(ABC)=$\frac{1}{2}AB.CP=500$ or, $10CP=500$ or,$CP=50$
We know, $\triangle ABC~\triangle CF'G'$
$\frac {AB}{F'G'}=\frac {CP}{CQ}=k$
So, $F'G'=\frac{AB}{k}=\frac{20}{k}$, $CQ=\frac{CP}{k}=\frac{50}{k}$
$(CF'G')=\frac{CQ.F'G'}{2}=\frac{20*50}{2.k.k}=\frac{500}{k.k}=80$ or,$k=\frac{5}{2}$
$FG=F'G'=\frac{20}{k}=8$, $CQ=\frac{50}{k}=20$
Then, $CP+C'P'=50+20=70$ or,$CQ+C'P'+PR+RQ=70$ or,$2CR=70$ or,$CR=35$
We proved, $\triangle CF'G'~\triangle CDE$
So, $\frac{DE}{F'G'}=\frac{CR}{CQ}=\frac{35}{20}=\frac{7}{4}$
$DE=\frac{7F'G'}{4}=\frac{7*8}{4}=14$
So, $(CDE)=\frac{1}{2}DE.CR=\frac{14*35}{2}=245$; that we need.

very easy problem