Question 19

Step 1 · Analyze the conditions on diagonals

When the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram.

When a parallelogram has equal diagonals, it is a rectangle (whose angles are each $90^\circ$).

A rectangle does not necessarily have all four sides equal. For it to be a square, its adjacent sides must also be equal or its diagonals must be perpendicular.

Since a rectangle is not always a square, the statement is not always true.

Step 1 · Find the measure of the fourth angle

Let the four angles of the quadrilateral be $\angle A, \angle B, \angle C,$ and $\angle D$.

Given that three angles are right angles: $\angle A = 90^\circ, \quad \angle B = 90^\circ, \quad \angle C = 90^\circ$

Using the angle sum property of a quadrilateral:

Since all four angles are $90^\circ$, opposite angles are equal, making it an equiangular parallelogram, which is a rectangle.

Step 1 · Prove opposite sides are parallel using triangle congruence

Let quadrilateral $\text{ABCD}$ have diagonals $\text{AC}$ and $\text{BD}$ intersecting at point $\text{O}$, such that $\text{AO} = \text{OC}$ and $\text{BO} = \text{OD}$.

In $\triangle \text{AOB}$ and $\triangle \text{COD}$:

• $\text{AO} = \text{OC}$ (Given)
• $\text{BO} = \text{OD}$ (Given)
• $\angle \text{AOB} = \angle \text{COD}$ (Vertically opposite angles)

By SAS congruence criterion: $\triangle \text{AOB} \cong \triangle \text{COD}$

Therefore, by CPCTC:

• $\text{AB} = \text{CD}$
• $\angle \text{OAB} = \angle \text{OCD}$ (Alternate interior angles $\implies \text{AB} \parallel \text{CD}$)

Similarly, in $\triangle \text{AOD}$ and $\triangle \text{COB}$:

• $\text{AO} = \text{OC}$
• $\text{DO} = \text{OB}$
• $\angle \text{AOD} = \angle \text{COB}$

By SAS congruence criterion: $\triangle \text{AOD} \cong \triangle \text{COB}$

Therefore, by CPCTC:

• $\text{AD} = \text{CB}$
• $\angle \text{ODA} = \angle \text{OBC}$ (Alternate interior angles $\implies \text{AD} \parallel \text{CB}$)

Since both pairs of opposite sides are parallel ($\text{AB} \parallel \text{CD}$ and $\text{AD} \parallel \text{CB}$), $\text{ABCD}$ is a parallelogram.

Step 1 · Check with counterexamples

A rhombus has perpendicular diagonals that bisect each other, and all four sides are equal.

However, having perpendicular diagonals alone does not guarantee all four sides are equal or that the diagonals bisect each other.

Counterexample: A kite has diagonals perpendicular to each other, but only two pairs of adjacent sides are equal, and all four sides are not equal. An arbitrary quadrilateral can also have perpendicular diagonals without being a rhombus.

Step 1 · Prove opposite sides are parallel using consecutive interior angles

Let quadrilateral $\text{ABCD}$ have $\angle A = \angle C$ and $\angle B = \angle D$.

Using the angle sum property of a quadrilateral: $\angle A + \angle B + \angle C + \angle D = 360^\circ$

Substitute $\angle C = \angle A$ and $\angle D = \angle B$:

Since consecutive interior angles $\angle A$ and $\angle B$ sum to $180^\circ$ for lines $\text{AD}$ and $\text{BC}$ with transversal $\text{AB}$: $\text{AD} \parallel \text{BC}$

Similarly:

Since both pairs of opposite sides are parallel, $\text{ABCD}$ is a parallelogram.

Step 1 · Calculate the value of each angle

Let each equal angle of the quadrilateral be $x$.

Using the angle sum property:

Since all four angles are $90^\circ$, opposite angles are equal (which makes it a parallelogram) and each angle is a right angle. Therefore, the quadrilateral is a rectangle.

Step 1 · Compare the definitions of trapezium and parallelogram

An isosceles trapezium is defined as a quadrilateral with exactly one pair of parallel opposite sides, while the non-parallel opposite sides are equal in length.

A parallelogram requires both pairs of opposite sides to be parallel.

Since an isosceles trapezium has only one pair of parallel sides, it cannot be a parallelogram.