Introduction
In trapezium ABCD, with AB parallel to CD, and AB equal to BC, we can derive several geometric properties.
1. Proving √A = √B
- Since AB || CD, angles A and B are equal (alternate interior angles).
- Triangle ABC is isosceles (AB = BC), making angles C and D equal.
- Hence, the areas A and B of triangles ABC and ABD correlate such that √A = √B.
2. Proving √C = √D
- Similar reasoning applies to triangles BCD and ACD.
- Both triangles share a common base CD and height from point A to CD.
- Thus, the areas C and D are proportional, leading to √C = √D.
3. Proving ∆ABC = ∆BAD
- The triangles share a common height from point A to line BC.
- The bases AB and BD are equal since AB = BC.
- Consequently, the areas of triangles ABC and BAD are equal, establishing ∆ABC = ∆BAD.
4. Proving diagonal AC = diagonal BD
- In trapezium ABCD, triangles ABC and ABD are both isosceles.
- Given that AB is parallel to CD and the angles at B and C are equal, sides AC and BD also become equal.
- Therefore, diagonals AC and BD are equal.
Conclusion
In summary, the properties of trapezium ABCD illustrate that √A = √B, √C = √D, ∆ABC = ∆BAD, and AC = BD, demonstrating the symmetry and congruence present in this geometric figure.