Abcd is a rhombus, eabf is a straight line such that ea=ab=bf. Prove that ed and fc when produced meet at right angle.?

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Answer

Understanding the Problem
We have a rhombus ABCD, and a straight line EABF where EA = AB = BF. We need to prove that the lines ED and FC, when produced, meet at a right angle.
Properties of Rhombus
- All sides of a rhombus are equal: AB = BC = CD = DA.
- The diagonals bisect each other at right angles.
Geometric Construction
- Place points E and F such that they lie on the straight line EABF.
- Given EA = AB = BF, we can denote the length of these segments as 'x'.
Key Angles and Triangles
- Since AB = EA, triangle EAB is isosceles with angles EAB = EBA.
- In rhombus ABCD, angle DAB = angle ABC (alternate angles created by line EA).
Finding Angles at Intersection Points
- When ED and FC are produced, we observe the angles formed.
- Since AB bisects the rhombus, angles DAB and ABC are equal.
Using Triangle Properties
- Triangles EAB and CBF are congruent due to:
- EA = AB
- AB = BF
- Common side AB.
- Therefore, angles EAB = FCB.
Conclusion
- Since angles DAB and ABC are equal, and EAB = FCB, we can infer that angles EDF and CFE must sum to 90 degrees.
- Thus, ED and FC, when produced, meet at right angles, concluding the proof.
This geometric relationship showcases the fascinating properties of rhombuses and congruent triangles, confirming that the lines ED and FC meet at a right angle under the given conditions.

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