Geometric Proofs

Proving the Angle Sum of a Triangle

• We can use parallel line rules to prove that the interior angles of a triangle sum to $180^\circ$.

• Step 1: Draw triangle $ABC$, then draw a straight line $XY$ passing through vertex $A$ that is exactly parallel to the base $BC$.

• Step 2: $\angle XAB = \angle ABC$ (Reason: Alternate angles are equal).

• Step 3: $\angle YAC = \angle ACB$ (Reason: Alternate angles are equal).

• Step 4: $\angle XAB + \angle BAC + \angle YAC = 180^\circ$ (Reason: Angles on a straight line sum to $180^\circ$).

• Step 5: Substitute the alternate angles to conclude $\angle ABC + \angle BAC + \angle ACB = 180^\circ$. (Hence proven).

Proving Triangle Congruence

• You can easily duplicate a digital shape perfectly, but how do we mathematically guarantee two drawn shapes are perfectly identical?
• Two shapes are congruent if they are identical in shape and size, regardless of any rotation, reflection, or translation.
• AQA requires you to use one of four specific criteria to prove triangle congruence: SSS (all three sides equal), SAS (two sides and the included angle), ASA (two angles and the included side), or RHS (Right-angle, hypotenuse, and one other side).
• Angle-Angle-Side (AAS) is also accepted because two equal angles guarantee the third is equal. However, you can NOT use AAA (which only proves similarity) or SSA (which does not guarantee a unique triangle).

Using Congruence in Practice

• Congruence proofs often require identifying hidden features, such as a common side that is structurally shared by two overlapping triangles.

• Worked Example: Prove $\triangle ABM \cong \triangle ACM$ where $M$ is the midpoint of the base $BC$ of an isosceles triangle $ABC$.

• Step 1: $AB = AC$ (Reason: Given, sides of an isosceles triangle are equal).

• Step 2: $BM = MC$ (Reason: $M$ is the midpoint of $BC$).

• Step 3: $AM = AM$ (Reason: Common side).

• Conclusion: Therefore, $\triangle ABM \cong \triangle ACM$ by SSS.

Geometric Similarity and Side Ratios

• Why does zooming in on a map keep the roads perfectly aligned, while stretching an image makes it look bizarrely distorted?
• Shapes are mathematically similar when they share identical angles but differ in size, meaning all corresponding side lengths are strictly proportional.
• To prove similarity in triangles, establishing that all corresponding angles are equal (AAA) is sufficient. Once established, you can calculate the scale factor ($k$) to deduce unknown side lengths.
• AQA requires a specific causal link in "Explain" questions. You must explicitly state: "Because the shapes are similar, all corresponding sides are in the same ratio."

Worked Example: Calculating Similar Lengths

• If triangle $ABC$ is similar to $DEF$, and you know $AB = 4$ cm, $DE = 10$ cm, and $BC = 6$ cm, you can find $EF$.

• Step 1: Identify corresponding sides ($AB$ corresponds to $DE$).

• Step 2: Calculate the scale factor by dividing the new length by the original length: $k = 10 \div 4 = 2.5$.

• Step 3: Multiply the corresponding known base by the scale factor: $EF = 6 \times 2.5 = 15$ cm.

• Alternatively, set up a ratio equation to show proportional equality, which is highly effective for "Show that" questions:

$$\frac{EF}{BC} = \frac{DE}{AB}$$

Pythagoras and Isosceles Proofs

• How can builders prove a corner is perfectly square without using a protractor?
• The converse of Pythagoras' theorem allows us to logically deduce right angles.
• Proof: Checking for Right Angles:
  • Step 1: Identify the longest side ($c$).
  • Step 2: Calculate $a^2 + b^2$ and $c^2$ separately.
  • Step 3: Concluding Statement: "Since $a^2 + b^2 = c^2$, the triangle is right-angled."
• Deductive reasoning also relies heavily on isosceles triangles, which feature two equal sides and two equal base angles.
• A common step in multi-stage proofs is applying the exterior angle theorem, which states the exterior angle of a triangle equals the sum of the two interior opposite angles.
• Proof: The Exterior Angle Theorem:
  • Step 1: Let a triangle have interior angles $a, b,$ and $c$. Let the exterior angle adjacent to $c$ be $d$.
  • Step 2: $a + b + c = 180^\circ$ (Reason: Angles in a triangle sum to $180^\circ$).
  • Step 3: $c + d = 180^\circ$ (Reason: Angles on a straight line sum to $180^\circ$).
  • Step 4: Therefore, $a + b + c = c + d$.
  • Step 5: Subtract $c$ from both sides: $a + b = d$. (Hence proven).

Deductive Properties of Quadrilaterals

• A square is actually a highly specific type of rhombus, kite, and parallelogram all rolled into one.
• Exam proofs frequently require explicit "Reasons" based on precise shape properties. For instance, the diagonals of a parallelogram bisect each other, while the diagonals of a rhombus and a kite intersect at exactly $90^\circ$.
• An isosceles trapezium is unique because its non-parallel legs are equal in length, and its base angles are identical.
• A cyclic quadrilateral requires all four vertices to sit exactly on the circumference of a circle; if even one vertex is at the centre, it does NOT have cyclic properties.

Proving Cyclic Quadrilateral Properties

• The defining geometric property of a cyclic quadrilateral is that its opposite angles sum to $180^\circ$.

• Step 1: In cyclic quadrilateral $ABCD$, draw radii from centre $O$ to vertices $B$ and $D$.

• Step 2: Let $\angle BAD = x$. The angle at the centre $\angle BOD = 2x$ (Reason: Angle at centre is twice angle at circumference).

• Step 3: Let $\angle BCD = y$. The reflex angle at centre $\angle BOD = 2y$.

• Step 4: $2x + 2y = 360^\circ$ (Reason: Angles around a point sum to $360^\circ$).

• Step 5: Divide by 2 to get $x + y = 180^\circ$. (Hence proven).