A line parallel to a side that halves the area of a triangle
A line parallel to a side that halves the area of a triangle
IP 2
7 marks
In triangle \(ABC\), \(D\) is a point on \(AB\) and \(E\) is a point on \(AC\) such that \(DE\) is parallel to \(BC\).
- Given that \(AD:DB=2:3\), find the ratio of the area of triangle \(ADE\) to the area of the trapezium \(DBCE\).[3]
- Instead, \(D\) is chosen so that the area of triangle \(ADE\) equals the area of the trapezium \(DBCE\). Find the exact value of \(\dfrac{AD}{AB}\), and hence show that \(AD:DB=(\sqrt{2}+1):1\).[4]
Solution:
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Answer:(a) \(4:21\) (b) \(\dfrac{AD}{AB}=\dfrac{\sqrt{2}}{2}\); \(AD:DB=(\sqrt{2}+1):1\)
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