The Construction Toolkit

Definition 1. A rhombus is a shape made up of four straight line segments whose side lengths are all equal.

In the rhombus ABCD above, we call AC and BD the diagonals of the rhombus that intersect at R.

Problem 1. Show that the diagonals of a rhombus are perpendicular bisectors of each other.

(Click for Solution)

Solution. Since angles in a triangle are supplementary,

$\begin{aligned} \angle ABD + \angle ADB + \angle BAD &= 180^\circ, \\ \angle CBD + \angle CDB + \angle BCD &= 180^\circ. \end{aligned}$

Since

$\begin{aligned} \angle ABC &= \angle ABD + \angle CBD,\\ \angle ADC &= \angle ADB + \angle CDB, \end{aligned}$

summing the equations yields

$\begin{aligned} \angle ABC + \angle ADC + \angle BAD + \angle BCD &= 360^\circ. \end{aligned}$

Since $BA = BC$ , $\Delta BAC$ is isosceles, so that $\angle BAC = \angle BCA$ . Similarly,

$\angle ABD = \angle ADB,\quad \angle CBD = \angle CDB, \quad \angle DAC = \angle DCA.$

Therefore,

$\begin{aligned} \angle ABC &= \angle ABD + \angle CBD \\ &= \angle ADB + \angle CDB \\ &= \angle ADC. \end{aligned}$

Similarly, $\angle BAD = \angle BCD$ . Hence,

$\begin{aligned} \angle ABC + \angle ADC + \angle BAD + \angle BCD &= 360^\circ \\ \angle ABC + \angle ABC + \angle BAD + \angle BAD &= 360^\circ \\ 2 \times (\angle ABC + \angle BAD) &= 360^\circ \\ \angle ABC + \angle BAD &= 180^\circ. \end{aligned}$

Since the interior angles sum to $180^\circ$ , $AD \parallel BC$ . By alternate angles,

$\angle ADB = \angle CBD = \angle CDB.$

Using previous data, we have $AD = CD$ ,

$\begin{aligned} \angle ADR &= \angle ADB = \angle CDB = \angle CDR, \\ \angle RAD &= \angle CAD = \angle ACD = \angle RCD. \end{aligned}$

By the ASA Criterion, $\Delta ADR \equiv \Delta CDR$ . Similarly,

$\Delta ADR \equiv \Delta CDR \equiv \Delta CBR \equiv \Delta ABR.$

Therefore, $AR = CR$ and $BR = DR$ . Furthermore, since adjacent angles on a straight line are supplementary,

$\begin{aligned} 2 \angle ARB &= \angle ARB + \angle ARB \\ &= \angle ARB + \angle CRB \\ &= 180^\circ, \end{aligned}$

so that $\angle ARB = 90^\circ$ , and similarly,

$\angle ARB = \angle CRB = \angle CRD = \angle ARD = 90^\circ.$

Therefore, $AC$ and $BD$ are perpendicular bisectors of each other.

Problem 2. Let AB be the line segment below, and let r > 0 be a length such that

AB/2 < r ≤ AB.

Here, C_A is a circle with centre A and radius r. Define C_B likewise. Let P, Q denote the intersections between C_A, C_B.

Show that the line passing through P, Q is the perpendicular bisector of AB.

(Click for Solution)

Solution. As radii of the same circle,

$AP = AQ = r = BP = BQ.$

Therefore, $APBQ$ forms a rhombus. By Problem 1, $PQ$ lies on the perpendicular bisector of $AB$ .

Problem 3. Let OA, OB be the line segments below, and let r > 0 be any length.

Here, C_O denotes the circle with centre O and radius r. Suppose C_O intersects OA at P and OB at Q. Define C_A, C_B similarly, and denote their intersection by R.