Question
Is it necessary to construct the pairs of arcs above and below XY? Instead, can we construct both pairs of arcs on the same side of XY? Explore this through construction, and then justify your answer.
✓
Answer
Draw a line segment XY.
Choose distances k and k’ which are slightly greater than half of the distance XY.
With centres at X and Y, draw arcs of radius ‘k’ above XY.
With centres at A and Y, draw arcs of radius ‘k’’ above XY.
Let the arcs intersect at the points A and B.
Join A and B and produce this line to intersect XY at O.
Join AX, AY, BX, and BY.
∆ABX and ∆ABY are congruent because AX = AY = k, BX = BY = k’, and AB is common.
∴ ∠XAO = ∠YAO
∆AOX and ∆AOY are congruent because AX = AY = k, ∠XAO = ∠YAO, and OA is common.
∴ OX = OY and ∠AOX = ∠AOY
Also, ∠AOX + ∠AOY = 180°
∴ 2∠AOX = 180° or ∠AOX = 90°
∴ OX = OY and ∠AOX = ∠AOY = 90°
∴ AB is the perpendicular bisector of the line XY.
Here, the pairs of arcs are both on the same side of XY.
∴ It is not necessary to construct the pairs of arcs above and below XY.
Choose distances k and k’ which are slightly greater than half of the distance XY.
With centres at X and Y, draw arcs of radius ‘k’ above XY.
With centres at A and Y, draw arcs of radius ‘k’’ above XY.
Let the arcs intersect at the points A and B.
Join A and B and produce this line to intersect XY at O.
Join AX, AY, BX, and BY.
∆ABX and ∆ABY are congruent because AX = AY = k, BX = BY = k’, and AB is common.
∴ ∠XAO = ∠YAO
∆AOX and ∆AOY are congruent because AX = AY = k, ∠XAO = ∠YAO, and OA is common.
∴ OX = OY and ∠AOX = ∠AOY
Also, ∠AOX + ∠AOY = 180°
∴ 2∠AOX = 180° or ∠AOX = 90°
∴ OX = OY and ∠AOX = ∠AOY = 90°
∴ AB is the perpendicular bisector of the line XY.
Here, the pairs of arcs are both on the same side of XY.
∴ It is not necessary to construct the pairs of arcs above and below XY.
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