4-gons (Quadrilaterals) - Prove key properties and characterisations of parallelograms

In a quadrilateral $ABCD$, $\angle A = 115^\circ$, $\angle B = 65^\circ$, and $\angle C = 115^\circ$. Determine if $ABCD$ is a parallelogram.

Step 1: Use the angle sum property of a quadrilateral to find the fourth angle $\angle D$. $\angle A + \angle B + \angle C + \angle D = 360^\circ$ Step 2: Substitute the known values into the equation: $115^\circ + 65^\circ + 115^\circ + \angle D = 360^\circ$ $295^\circ + \angle D = 360^\circ$ Step 3: Solve for $\angle D$: $\angle D = 360^\circ - 295^\circ = 65^\circ$ Step 4: Check the pairs of opposite angles. $\angle A = 115^\circ$ and $\angle C = 115^\circ$ (Equal). $\angle B = 65^\circ$ and $\angle D = 65^\circ$ (Equal). Since both pairs of opposite angles are equal, $ABCD$ is a parallelogram.

This solution applies the theorem that a quadrilateral is a parallelogram if its opposite angles are equal. We first calculate the missing angle to verify the condition for both pairs.

In quadrilateral $PQRS$, diagonals $PR$ and $QS$ intersect at $O$. Given $PO = x + 2$, $OR = 10$, $QO = y - 3$, and $OS = 12$. Find the values of $x$ and $y$ that make $PQRS$ a parallelogram.

Step 1: For $PQRS$ to be a parallelogram, the diagonals must bisect each other. This means $PO = OR$ and $QO = OS$. Step 2: Set up the equation for diagonal $PR$: $x + 2 = 10$ $x = 10 - 2 = 8$ Step 3: Set up the equation for diagonal $QS$: $y - 3 = 12$ $y = 12 + 3 = 15$ Step 4: Therefore, for $x = 8$ and $y = 15$, the diagonals bisect each other, making $PQRS$ a parallelogram.

This example uses the diagonal bisection property. By equating the two segments of each diagonal, we ensure the intersection point is the midpoint for both, which is a necessary and sufficient condition for a parallelogram.

1. Since one pair of opposite sides ($AB$ and $CD$) are both equal and parallel, $ABCD$ is a parallelogram.
2. In a parallelogram, opposite angles are equal.
3. Therefore, $\angle ABC = \angle ADC$.
4. Given $\angle ADC = 70^\circ$, so $\angle ABC = 70^\circ$.

This uses the property that if one pair of opposite sides is equal and parallel, the quadrilateral must be a parallelogram, and hence opposite angles are equal.

1. In a parallelogram, opposite sides are equal. Therefore, $PQ = SR$.
2. $3x - 5 = x + 7$
3. $3x - x = 7 + 5$
4. $2x = 12$
5. $x = 6$
6. $PQ = 3(6) - 5 = 18 - 5 = 13$ cm.

We equate the expressions for opposite sides because opposite sides of a parallelogram are always equal in length.