Question 980769
in a triangle ABC, D and E are two points on sides AB and AC respectively so
that DE parallel to BC and AD/DB=2/3.Then, the area of trapezium DECB/the area
of triangle ABC is equal to ?
<pre>
{{{drawing(400,560/3,-4,11,-1,6,triangle(-3,0,10,0,0,5),
triangle(-6/5,3,4,3,0,5),
locate(-3,0,B), locate(10,0,C),locate(-1.6,3.3,D), locate(4.2,3.4,E),
locate(-.1,5.5,A)


  )}}}

We will use three theorems and the fact that &#916;ADE&#8765;&#916;ABC.     :

(1).  If {{{p/q=r/s}}}, then {{{p/(p+q)=r/(r+s)}}}

(2).  If {{{p/q=r/s}}}, then {{{(q-p)/q=(s-r)/s}}}

(3).  In two similar triangles, the ratio of their areas is the square
 of the ratio of their sides.

{{{AD/DB=2/3}}}  given

{{{AD/(AD+DB)=2/(2+3)}}} using (1)

{{{AD/AB=2/5}}} simplifying

{{{matrix(1,4,"(area",of,triangle,"ADE)")/matrix(1,4,"(area",of,triangle,"ABC)")=4/25}}} using (3)

{{{(matrix(1,4,"(area",of,triangle,"ABC)")-matrix(1,4,"(area",of,triangle,"ADE)"))/matrix(1,4,"(area",of,triangle,"ABC)")=(25-4)/25}}} using (2)

{{{matrix(1,4,"(area",of,trapezium,"DECB)")/matrix(1,4,"(area",of,triangle,"ABC)")=21/25}}}, simplifying

Edwin</pre>