Prove that the diagonals of a rectangle bisect each other at right angle. (using congruency)
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Prove that the diagonals of a rectangle bisect each other at right angle. (using congruency)
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Answer:
To prove that the diagonals of a rectangle bisect each other at a right angle, we can use the concept of congruent triangles. Let's consider a rectangle ABCD.
In a rectangle:
1. Opposite sides are equal in length.
2. All angles are right angles.
Now, let's consider the diagonals AC and BD of the rectangle ABCD.
To prove that these diagonals bisect each other at a right angle, we can show that the triangles formed by the diagonals and the midpoints of the opposite sides are congruent.
Consider the triangles:
1. △ABC (with AC as the base) and △ADC (with DC as the base).
2. △ABD (with AB as the base) and △CBD (with CB as the base).
Using the properties of a rectangle:
1. AB = CD (opposite sides of a rectangle are equal)
2. BC = AD (opposite sides of a rectangle are equal)
3. ∠B = ∠D = 90° (all angles of a rectangle are right angles)
Now, consider the midpoints of the sides AB, BC, CD, and AD. Let E be the midpoint of AB, F be the midpoint of BC, G be the midpoint of CD, and H be the midpoint of AD.
Using the midpoint property, we can say that:
1. AE = EB
2. BF = FC
3. CG = GD
4. AH = HD
Now, we can prove the triangles congruent using the following criteria:
1. △AED ≅ △CEB (Side-Angle-Side congruence: AE = EB, ∠E = ∠B = 90°, AB = CD)
2. △BFC ≅ △DFC (Side-Angle-Side congruence: BF = FC, ∠F = ∠C = 90°, BC = AD)
Now, we have proved that the diagonals of the rectangle bisect each other (AE = EB and BF = FC) and they intersect at right angles (∠E = ∠B = ∠F = ∠C = 90°). Hence, the diagonals AC and BD bisect each other at right angles in the rectangle ABCD.
Step-by-step explanation: