Geometry - Constructions & Proofs

The Logical Hierarchy

Geometric knowledge follows a strict hierarchy. An axiom is a statement we accept as true without proof, used as a starting point for argument. We use five: the Two Points Axiom, the Ruler Axiom, the Protractor Axiom, SAS+ASA+SSS (congruence), and the Axiom of Parallels. A theorem is a statement proven true through logical steps, using only axioms and results that come before it in the course sequence. A corollary is a short extra result that follows immediately from a theorem. If you use a later theorem, you may be quietly assuming the very thing you are trying to prove.

Formal Proofs: Theorems 11, 12, and 13

Theorems 11, 12, and 13 are the three proofs the syllabus names as ones you may be asked to reproduce at Higher Level. Junior Certificate Higher Level material is assumed knowledge, so keep the standard arguments (angle sum of a triangle, alternate angles, Pythagoras) at your fingertips too.

Theorem 11. If three parallel lines cut off equal segments on some transversal line, then they will cut off equal segments on any other transversal.

Given: Three parallel lines $AD \parallel BE \parallel CF$, and a transversal line $AC$ cutting them at $A$, $B$, and $C$ with $|AB| = |BC|$. A second transversal line $DF$ cuts the parallel lines at $D$, $E$, and $F$.

To prove: $|DE| = |EF|$.

Construction: Through $B$, draw a line parallel to $DF$, meeting $AD$ at $X$ and $CF$ at $Y$.

Proof:

• Statement: $DEBX$ is a parallelogram. Reason: $AD \parallel BE$ (given) and $BX \parallel DF$ (by construction).
• Statement: $|DE| = |BX|$. Reason: Opposite sides of a parallelogram are equal (Theorem 9).
• Statement: $EFYB$ is a parallelogram. Reason: $BE \parallel CF$ (given) and $BY \parallel DF$ (by construction).
• Statement: $|EF| = |BY|$. Reason: Opposite sides of a parallelogram are equal (Theorem 9).
• Statement: $|AB| = |CB|$. Reason: Given ($|AB| = |BC|$).
• Statement: $|\angle XAB| = |\angle YCB|$. Reason: Alternate angles ($AD \parallel CF$).
• Statement: $|\angle ABX| = |\angle CBY|$. Reason: Vertically opposite angles.
• Statement: $\triangle ABX \cong \triangle CBY$. Reason: ASA.
• Statement: $|BX| = |BY|$. Reason: Corresponding sides in congruent triangles.
• Statement: $|DE| = |EF|$. Reason: Since $|DE| = |BX|$ and $|EF| = |BY|$. Q.E.D.

Theorem 12. Let $\triangle ABC$ be a triangle. If a line is parallel to $BC$ and cuts $[AB]$ in the ratio $s : t$, then it cuts $[AC]$ in the same ratio.

A point $P$ divides $[AB]$ in the ratio $s : t$ if it lies on $AB$ between $A$ and $B$ with $|AP| / |PB| = s / t$ — so 'cuts $[AB]$ in the ratio $2 : 3$' means the point sits two-fifths of the way from $A$ to $B$.

Given: $\triangle ABC$, and a line $l$ parallel to $BC$ cutting $[AB]$ at $D$ in the ratio $m : n$, with $m, n \in \mathbb{N}$ (the commensurable case).

To prove: $l$ cuts $[AC]$ in the ratio $m : n$.

Construction: Mark points along $[AB]$ dividing it into $m + n$ equal segments, with $D$ the $m$-th point, and through each point draw a line parallel to $BC$.

Proof:

• Statement: These parallel lines cut $[AC]$ into $m + n$ equal segments. Reason: Theorem 11.
• Statement: There is only one line through $D$ parallel to $BC$, so $l$ is that line and meets $[AC]$ at the $m$-th point $E$. Reason: Axiom of Parallels.
• Statement: $E$ divides $[AC]$ in the ratio $m : n$ (i.e. $|AE| / |EC| = m / n$). Reason: $[AE]$ contains $m$ of the equal segments and $[EC]$ contains $n$ of them. Q.E.D.

Theorem 13. If two triangles $\triangle ABC$ and $\triangle A'B'C'$ are similar, then their sides are proportional: $\frac{|AB|}{|A'B'|} = \frac{|BC|}{|B'C'|} = \frac{|CA|}{|C'A'|}$.

Given: Similar triangles $\triangle ABC$ and $\triangle A'B'C'$ with equal corresponding angles: $|\angle A| = |\angle A'|$, $|\angle B| = |\angle B'|$, and $|\angle C| = |\angle C'|$. Assume $|A'B'| \le |AB|$.

To prove: $\frac{|AB|}{|A'B'|} = \frac{|BC|}{|B'C'|} = \frac{|CA|}{|C'A'|}$.

Construction: Mark $B''$ on $[AB]$ with $|AB''| = |A'B'|$ and $C''$ on $[AC]$ with $|AC''| = |A'C'|$. Join $B''C''$.

Proof:

• Statement: $\triangle AB''C'' \cong \triangle A'B'C'$. Reason: SAS ($|AB''| = |A'B'|$, $|\angle A| = |\angle A'|$, $|AC''| = |A'C'|$).
• Statement: $|\angle AB''C''| = |\angle B'| = |\angle ABC|$. Reason: Corresponding angles in congruent triangles.
• Statement: $B''C'' \parallel BC$. Reason: Corresponding angles are equal.
• Statement: $\frac{|AB|}{|AB''|} = \frac{|AC|}{|AC''|}$, which rearranges to $\frac{|AB|}{|A'B'|} = \frac{|AC|}{|A'C'|}$. Reason: Theorem 12, since $B''C'' \parallel BC$.
• The same argument starting from vertex $B$ gives $\frac{|BC|}{|B'C'|} = \frac{|AB|}{|A'B'|}$.
• Therefore, $\frac{|AB|}{|A'B'|} = \frac{|BC|}{|B'C'|} = \frac{|CA|}{|C'A'|}$. Q.E.D.

Congruence and Similarity

You are often asked to prove triangles are congruent. Use only: SAS (two sides and the included angle), ASA (two angles and a side), SSS (three sides), or RHS (right-angled triangles: hypotenuse and one other side). Two triangles are similar if all three angles are equal, or if one angle is equal and the sides around it are in proportion. When writing proofs, use a two-column (Statement | Reason) format. Order the vertices so they correspond: if △ACD ≅ △BCE, A must match B, C must match C, and D must match E. Marks are lost if the order is wrong.

Constructions 16–22

Leaving Certificate Higher Level requires knowledge of Constructions 16–22:

• 16. Circumcentre and circumcircle (compass and straight-edge only): Construct the perpendicular bisectors of any two sides of the triangle. Their intersection is the circumcentre. Place the compass at this point, set the radius to any vertex, and draw the circumcircle.
• 17. Incentre and incircle (compass and straight-edge only): Construct the internal bisectors of any two angles of the triangle. Their intersection is the incentre. Construct a perpendicular line from the incentre to one side to find the inradius, then draw the incircle inside the triangle.
• 18. Angle of 60° (without protractor or set square): With the same compass opening, draw an arc from vertex $A$ cutting ray $[AB$ at $P$, and from $P$ draw an arc crossing the first at $Q$. Joining $A$ through $Q$ forms an equilateral triangle, so $|\angle QAB| = 60^\circ$.
• 19. Tangent to a given circle at a given point on it: Draw the radius $[OP]$ from the centre $O$ to point $P$ on the circle. Construct the line through $P$ perpendicular to $OP$ (Theorem 20).
• 20. Parallelogram (given side lengths and an angle): Draw the base of given length, construct the given angle at one endpoint, mark off the second side length, and construct parallel lines or use compass arcs of the side lengths to locate the fourth vertex.
• 21. Centroid of a triangle: Find the midpoints of two sides (using perpendicular bisectors) and join each midpoint to the opposite vertex. The two medians meet at the centroid.
• 22. Orthocentre of a triangle: Construct the perpendicular (altitude) from one vertex to the opposite side, then repeat from a second vertex. Two altitudes are enough—they meet at the orthocentre. If the triangle is obtuse, extend the sides to drop your perpendiculars; the orthocentre lies outside the triangle.

Do not rub out your arcs; they are primary evidence that you used the correct tools.

Key terms

Axiom

A statement accepted as true without proof, used as a starting point for all geometric reasoning.

Converse

A statement formed by swapping the 'if' and 'then' clauses of a theorem. It must be verified independently.

Proof by Contradiction

Assuming the opposite of your target statement and showing it leads to an impossible outcome.

Circumcentre

The intersection of the perpendicular bisectors of the sides of a triangle; the centre of the circumcircle.

Incentre

The intersection of the internal angle bisectors of a triangle; the centre of the incircle.

Centroid

The intersection of the three medians of a triangle.

Orthocentre

The intersection of the three altitudes of a triangle.

If and only if

A condition meaning that P implies Q and Q implies P; you must provide two separate proofs.