Question 2

We can construct a perpendicular bisector using arcs on only one side.

Step 1 — Constructing the perpendicular bisector

Draw a line segment. Call it XY.

Open your compass. Make it wider than half of XY. Keep this opening. Let us call it k. Put the compass point on X. Draw an arc above XY. Put the compass point on Y. Draw another arc above XY. These two arcs meet. Call this point A.

Now, change your compass opening. Make it another width. Ensure this new width is also greater than half of XY. Let us call this new opening k'. Put the compass point on X. Draw an arc above XY. Put the compass point on Y. Draw another arc above XY. These two new arcs meet. Call this point B.

Draw a straight line. Connect point A to point B. Extend this line. It cuts XY at point O. Join AX, AY, BX, and BY with straight lines.

Step 2 — Justifying the construction

Consider two large triangles. These are $\triangle ABX$ and $\triangle ABY$. From our construction, $AX = AY$. This is because we used the same radius k. Also, $BX = BY$. This is because we used the same radius k'. The side $AB$ is common to both triangles. So, $\triangle ABX$ is congruent to $\triangle ABY$. This is by the SSS congruence rule. Congruent triangles have equal corresponding parts. So, $\angle XAB$ is equal to $\angle YAB$.

Now, consider two smaller triangles. These are $\triangle AOX$ and $\triangle AOY$. We know $AX = AY$ (radius k). We just showed $\angle XAO = \angle YAO$. The side $AO$ is common to both triangles. So, $\triangle AOX$ is congruent to $\triangle AOY$. This is by the SAS congruence rule. Again, corresponding parts of congruent triangles are equal. So, $OX = OY$. This means O is the midpoint of XY. Also, $\angle AOX = \angle AOY$. Angles $\angle AOX$ and $\angle AOY$ form a straight line. Their sum is $\mathbf{180}$ degrees.

$\angle AOX + \angle AOY = 180^\circ$

Since they are equal, we can write:

$2 \times \angle AOX = 180^\circ$

Divide both sides by 2.

$\angle AOX = \frac{180^\circ}{2}$

$\boxed{\angle AOX = 90^\circ}$

So, the line AB is perpendicular to XY. Line AB bisects XY at O. It is also perpendicular. Therefore, AB is the perpendicular bisector of XY.

Answer

It is not necessary to construct arcs above and below XY. Arcs on the same side of XY are enough.