Given a rectangle ABCD. A diagonal is drawn from vertex A. From the vertex D to the side BC, a line is drawn (to point E), which intersects the diagonal at point O, and EO is two times less than OD. Find the area of the triangle COE if the area of the rectangle is ABCD 60.
After the construction, we get triangles, compare the AOD and COE triangles:
<EOC = <AOD – as vertical angles at the intersection of AC and DE.
<ECO = <DAO – as internal criss-crossing.
<ADO = <OEC – as remaining.
It turns out that the AOD and COE triangles are similar in three corners.
In such triangles, the corresponding sides are proportional. We have OD / EO = 2
EO / OD = AD / EC = Ao / OE
AD = 2EC,
Let’s construct the heights in these triangles from point O: OZ – triangle AOD, OK – triangle COE.
As you can see, ZO + OK = DC = AB
ZO / OK = 2, similar to triangles.
ZO = 2OK
AB = 2OK + OK = 3OK
Expression for determining the area ABCD:
S = AB * AD = 60
S = 3 * OK * 2 * EC = 60
OK * EC = 60/6 = 10
Expression for determining the area of the COE triangle:
S = (1/2) * EC * OK = (1/2) * 10 = 5
Answer: The area of the СOE triangle is 5.
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