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Introduction
In trapezium ABCD, where BD and AC are diagonals intersecting at point O, we are to demonstrate that the point of trisection of both diagonals is established under the condition that BC = 2AD.
Understanding the Configuration
- Let AD be the shorter base and BC the longer base of the trapezium.
- Define the lengths: AD = x and BC = 2x.
Using Similar Triangles
- Triangles AOD and BOC share a common angle at O.
- Since AD is parallel to BC, angles AOD and BOC are equal.
- This makes triangles AOD and BOC similar by the Angle-Angle similarity criterion.
Establishing Ratios
- From the similarity of triangles, we have the ratio:
AO/BO = AD/BC = x/(2x) = 1/2.
- Therefore, AO:BO = 1:2.
Finding Points of Trisection
- Let the lengths of diagonals AC and BD be divided into three equal parts at points P and Q respectively.
- Given that AO:OB = 1:2, point O divides diagonal AC in the ratio 1:2.
- Now considering diagonal BD, with the same ratio, point O divides BD in a similar fashion.
Conclusion
- Thus, we can conclude that the points of trisection of both diagonals AC and BD are at the same relative positions, confirming that O is indeed a point of trisection as required.
- The key takeaway is that, in trapezium ABCD with the condition BC = 2AD, the diagonals intersect at points that trisect their lengths proportionally.