Geometry of Triangles and Quadrilaterals for the ESAT

Understanding Geometric Properties

To solve complex geometry problems in the ESAT, you must be able to apply the fundamental properties of triangles and quadrilaterals simultaneously. This often involves looking for hidden relationships, such as shared sides that create new isosceles triangles or provide enough information to prove congruence between two shapes. You should be familiar with the properties of isosceles and equilateral triangles, as well as the defining characteristics of rectangles, parallelograms, and kites.

Properties of Isosceles Triangles

Isosceles triangles are a frequent component of geometric diagrams. When two isosceles triangles share a side, it often allows you to equate lengths across the entire figure.

Consider the following example. In the isosceles triangle $ABC$, $AB = AC$ and angle $BAC = 50^\circ$. Another isosceles triangle $CAD$ is drawn in the same plane as triangle $ABC$, with angle $CAD = 18^\circ$.

To find the size of angle $DBC$, we first mark all equal lengths onto the diagram: $AB = AC = AD$. This shows that triangle $ABD$ is also isosceles, as it has two sides of equal length ($AB = AD$).

Next, we join $B$ and $D$ to form the triangle $DBC$, which contains our target angle.

1. Calculate the base angles of $\triangle ABC$: $\text{angle } ABC = \text{angle } ACB = \frac{180 - 50}{2} = 65^\circ$ because base angles of an isosceles triangle are equal.
2. Determine the total angle at $A$: $\text{angle } BAD = 50^\circ + 18^\circ = 68^\circ$.
3. Calculate the base angles of the new isosceles $\triangle ABD$: $\text{angle } ABD = \text{angle } ADB = \frac{180 - 68}{2} = 56^\circ$.
4. Find the final difference: $\text{angle } DBC = 65^\circ - 56^\circ = 9^\circ$.

Combining Triangles and Special Quadrilaterals

Often, quadrilaterals and triangles are combined to test your knowledge of symmetry and congruence. In the following diagram, $PQRS$ is a rectangle while $PQT$ and $QRU$ are equilateral triangles.

If the length of $TS$ is 10 cm, what is the length of $TU$? To solve this, we must identify congruent triangles by marking equal sides:

• $SP = QR$ (opposite sides of a rectangle are equal).
• $QR = RU = QU$ (sides of equilateral triangle $QRU$ are equal).
• $SR = PQ = PT = QT$ (opposite sides of a rectangle and sides of equilateral triangle $PQT$).

In triangles $SPT$ and $UQT$:

1. $SP = QU$ (since both equal $QR$).
2. $PT = QT$ (since both are sides of equilateral $\triangle PQT$).

Now we check the included angle between these sides:

• $\text{angle } SPT = 90^\circ + 60^\circ = 150^\circ$.
• $\text{angle } TQU = 360^\circ - (60^\circ + 90^\circ + 60^\circ) = 360^\circ - 210^\circ = 150^\circ$ (using the angles around the point $Q$).

Because they share two sides and an identical included angle, triangle $SPT$ is congruent to triangle $UQT$ (SAS). This means the triangles are identical in every way, so $TU = TS = 10$ cm.

Applying Parallelogram and Kite Properties

Parallelograms and kites provide useful angle facts based on symmetry. Opposite angles of a parallelogram are equal, and kites have one pair of equal opposite angles along their axis of symmetry.

Example: Parallelogram and Isosceles Triangles

$ABCD$ is a parallelogram. $CDE$ is an isosceles triangle with $CD = CE$, and $ADE$ is a straight line. Angle $CED = 65^\circ$. Triangle $BFC$ is isosceles with $BF = FC$ and angle $BFC = 110^\circ$.

To find the size of angle $ABF$:

1. Find angles in $\triangle CDE$: $\text{angle } CDE = 65^\circ$ (base angles of an isosceles triangle).
2. Use the straight line property: $\text{angle } ADC = 180^\circ - 65^\circ = 115^\circ$.
3. Use parallelogram properties: $\text{angle } ABC = \text{angle } ADC = 115^\circ$ (opposite angles are equal).
4. In $\triangle FBC$, the base angles are $\text{angle } FBC = \frac{180 - 110}{2} = 35^\circ$.

Finally, $\text{angle } ABF = \text{angle } FBC + \text{angle } ABC = 35^\circ + 115^\circ = 150^\circ$.

Example: Kite and Isosceles Triangles

$ABCD$ is a kite. $ABE$ is an isosceles triangle with $EA = EB$. Angle $ADC = 105^\circ$. Angle $DCB = 24^\circ$ and angle $AEB = 30^\circ$.

To find angle $DBE$:

1. Identify symmetry: $\text{angle } ABC = \text{angle } ADC = 105^\circ$.
2. Find total internal angles: $\text{angle } DAB = 360^\circ - (105^\circ + 105^\circ + 24^\circ) = 126^\circ$.
3. Use isosceles $\triangle ABD$ ($AD = AB$ for a kite): $\text{angle } DBA = \frac{180 - 126}{2} = 27^\circ$.
4. Use isosceles $\triangle ABE$: $\text{angle } EBA = \frac{180 - 30}{2} = 75^\circ$.

Finally, $\text{angle } DBE = \text{angle } DBA + \text{angle } ABE = 27^\circ + 75^\circ = 102^\circ$.