Using straight edge and compasses only construct:?

Question

the perpendicular bisector of a line segment and explain the geometrical principles that ensure your constructions work.

Answer 1

Let the given line segment be AB.

With A as center and radius more than half the length of AB, draw an arc on each side of AB. With B as center and the same radius, draw an arc on each side of AB to cut the previously drawn arcs, say at X and Y.

Join XY.

Thus, XY is the perpendicular bisector of AB.

Proof :

Let XY and AB intersect at M.

In triangles AXY and BXY,
XY = XY (Common)
AX = BX (Radii of the same circle)
AY = BY (Radii of the same circle)
So AXY is congruent to BXY.

Thus, AXM = BXM (CPCT)

In triangles AXM and BXM,
AX = BX
XM =XM
AXM = BXM
So AXM congruent to BXM

AM = BM (CPCT)
Hence, XY is the bisector of AB.

AMX = BMX (CPCT)
AMX + BMX = 180 (Linear Pair)
Solving : AMX = BMX = 90
Hence, XY is perpendicular to AB

Hence proved.

Hope this helped.

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Answer 2

From each endpoint of the line strike arcs of equal radii, making sure your radius is larger than half the line length. Connect the intersections of the arcs with a straight line. This line bisects the original line segment. To prove this, connect the arc intersections to the original endpoints and note that you have produced 4 congruent triangles. Since the respective sides of the triangles have equal lengths, you have bisected the original line.