Congruent Triangles (GCSE Maths)

• Both have a right angle, as indicated by the red square symbol.
• The hypotenuse (the longest side opposite the right angle) is $10\text{ cm}$ in both triangles.
• One of the other corresponding legs is $6\text{ cm}$ in both triangles.

Having a right angle, equal hypotenuses, and equal corresponding sides satisfies the RHS condition.

• Check the measurements given for both triangles and ensure the units match.
• Convert the measurements of the second triangle from metres to centimetres:

Observe that both triangles have two sides measuring $120\text{ cm}$ and $80\text{ cm}$, and the angle strictly between these two sides (the included angle) is $60^\circ$ in both cases.

Since two sides and the included angle are equal, the condition is SAS.

The lengths of all three sides are provided for both shapes.

The sides of the first triangle ($9\text{ cm}$, $9.3\text{ cm}$, $9.5\text{ cm}$) perfectly match the corresponding sides of the second triangle.

Because all three corresponding sides are equal in length, the congruence condition is SSS.

• Both triangles contain a right angle ($90^\circ$).
• The longest side opposite the right angle (the hypotenuse) is given as $5\text{ cm}$ in both triangles.
• Another corresponding side is given as $3\text{ cm}$ in both triangles.

Since we have a right angle, equal hypotenuses, and one other equal side, the congruence condition is:

First, draw a diagonal line connecting vertex $P$ to vertex $R$ to split the quadrilateral into two triangles: triangle $PQR$ and triangle $PSR$.

Compare the sides of triangle $PQR$ and triangle $PSR$:

• We are given that the top two sides are equal: $PQ = PS$.
• We are given that the bottom two sides are equal: $QR = SR$.
• The diagonal line is shared by both triangles, meaning it is a common side: $PR = PR$.

Since all three corresponding pairs of sides are equal in length, the two triangles are congruent by the Side-Side-Side (SSS) condition:

Because the triangles are exactly congruent, all of their corresponding angles must also be equal. Therefore:

Because triangle $PQR$ is an equilateral triangle, all its sides are equal in length. This gives us our equal hypotenuses for the two smaller triangles:

Both triangles share the vertical line segment as a common side:

We are given that $PS$ is perpendicular to $RQ$, meaning both adjacent angles at $S$ are right angles:

• Since we have a right angle, equal hypotenuses, and one other equal corresponding side, the triangles are congruent by the Right-angle-Hypotenuse-Side (RHS) condition.
• Note: This can also be proven using the Angle-Angle-Side (AAS) condition by noting that $\angle PRS = \angle PQS = 60^\circ$ because triangle $PQR$ is equilateral.

Find the missing third angle in the first triangle using the fact that angles in a triangle sum to $180^\circ$:

• Looking at the first triangle, the side measuring $10\text{ cm}$ is situated exactly between the $19^\circ$ angle and the calculated $21^\circ$ angle.
• Looking at the second triangle, it also features a side of $10\text{ cm}$ situated exactly between a $19^\circ$ angle and a $21^\circ$ angle.
• Since two angles and the included side are equal, the condition is ASA. (Note: AAS is also an acceptable answer if comparing the $140^\circ$ and $19^\circ$ angles directly).

To prove that triangle $PST$ is congruent to triangle $RQT$, we use the properties of a parallelogram. Opposite sides of a parallelogram are equal in length:

Because the opposite sides are parallel ($PS \parallel RQ$) and $QS$ acts as a transversal line, the alternate interior angles are equal:

Similarly, using the transversal line $PR$, the other pair of alternate interior angles are also equal:

• Since we have two angles and the included side of one triangle equal to the corresponding parts of the other, the triangles are congruent by the Angle-Side-Angle (ASA) condition.
• Note: You can also use the Angle-Angle-Side (AAS) condition by identifying the vertically opposite angles $\angle PTS = \angle RTQ$.

First, calculate the missing angle in triangle $ABD$ using the property that angles in a triangle sum to $180^\circ$:

Angle: From the given diagram, $\angle ABD = 61^\circ$ and $\angle CDB = 61^\circ$.

Side: The line segment $BD$ is shared by both triangles.

Angle: From our calculation in step 1 and the given diagram, $\angle ADB = 54^\circ$ and $\angle CBD = 54^\circ$.

• Since two angles and the included side are equal, the triangles are congruent by the Angle-Side-Angle (ASA) condition.
• Note: You can also prove this using the Angle-Angle-Side (AAS) condition by calculating $\angle BCD = 65^\circ$ and comparing it to $\angle DAB = 65^\circ$.

The opposite sides of a rectangle are equal in length, therefore:

Similarly, the other pair of opposite sides are also equal:

Both triangles share the diagonal as a common side:

• Since all three pairs of corresponding sides are equal, the triangles are congruent by SSS.
• Note: You can also prove this using the Side-Angle-Side (SAS) or Right-angle-Hypotenuse-Side (RHS) conditions by stating that $\angle ABC = \angle ADC = 90^\circ$.