Strand 2 — Geometry & Trigonometry
Synthetic Geometry: Theorems & Constructions (HL)
5th Year · 6th Year (Leaving Cert)
- ✓By the end of this lesson students will be able to state and apply key theorems related to lines, triangles, and circles.
- ✓By the end of this lesson students will be able to understand and reproduce the proofs of Theorems 11, 12, and 13.
- ✓By the end of this lesson students will be able to perform geometric constructions using only a compass and straightedge.
- ✓By the end of this lesson students will be able to solve problems involving synthetic geometry, justifying steps with appropriate theorems.
Key concepts
Theorem 4: Vertically Opposite Angles
When two straight lines intersect, the angles opposite each other at the point of intersection are equal in measure.
Theorem 6: Sum of Angles in a Triangle
The sum of the measures of the three interior angles of any triangle is always 180 degrees.
Theorem 9: Isosceles Triangle Property
If two sides of a triangle are equal in length, then the angles opposite these sides are equal in measure.
Theorem 11 (Proof HL): Parallel Lines and Transversals
If three parallel lines cut off equal segments on some transversal line, then they will cut off equal segments on any other transversal line.
Proof:
1. Given: Three parallel lines l1, l2, l3. A transversal t1 intersects l1, l2, l3 at A, B, C respectively, such that |AB| = |BC|. Another transversal t2 intersects l1, l2, l3 at D, E, F respectively.
2. To Prove: |DE| = |EF|.
3. Construction: Through D, draw a line parallel to t1, meeting l2 at G and l3 at H. Through E, draw a line parallel to t1, meeting l3 at K.
4. Proof:
* Consider the quadrilateral ABGD. Since AB || DG (l1 || l2) and AD || BG (by construction, DG is parallel to t1), ABGD is a parallelogram. Therefore, |AB| = |DG|.
* Consider the quadrilateral BCEK. Since BC || EK (l2 || l3) and BE || CK (by construction, EK is parallel to t1), BCEK is a parallelogram. Therefore, |BC| = |EK|.
* Given |AB| = |BC|, it follows that |DG| = |EK|.
* Now consider triangles DGE and EKF:
* ∠GDE = ∠KEF (Corresponding angles, since DG || EK, both parallel to t1).
* ∠DEG = ∠EFK (Corresponding angles, since l2 || l3).
* Therefore, ΔDGE is equiangular to ΔEKF.
* Since ΔDGE and ΔEKF are equiangular and have a corresponding side equal (|DG| = |EK|), they are congruent (AAS).
* Thus, |DE| = |EF| (Corresponding sides of congruent triangles).
Theorem 12 (Proof HL): Proportional Division of Sides in a Triangle
Let ABC be a triangle. If a line l is parallel to BC and cuts [AB] in P and [AC] in Q, then |AB|/|AP| = |AC|/|AQ|.
Proof (for commensurable segments):
1. Given: ΔABC, line PQ || BC, P on [AB], Q on [AC].
2. To Prove: |AB|/|AP| = |AC|/|AQ|.
3. Assume |AP| and |PB| are commensurable. Let |AP|/|PB| = m/n, where m, n are positive integers.
4. Construction: Divide [AP] into m equal segments and [PB] into n equal segments. All these (m+n) segments are of equal length. Through the division points, draw lines parallel to BC.
5. Proof:
* By Theorem 11, since the parallel lines cut off equal segments on [AB], they also cut off equal segments on [AC].
* Therefore, [AQ] is divided into m equal segments and [QC] is divided into n equal segments.
* This implies |AQ|/|QC| = m/n.
* So, we have |AP|/|PB| = |AQ|/|QC|.
* From this, we can derive the required ratio:
* |PB|/|AP| = |QC|/|AQ|
* 1 + |PB|/|AP| = 1 + |QC|/|AQ|
* (|AP| + |PB|)/|AP| = (|AQ| + |QC|)/|AQ|
* |AB|/|AP| = |AC|/|AQ|.
(Note: The case where |AP| and |PB| are incommensurable is proved by a limiting argument, which is beyond the scope of this course.)
Theorem 13 (Proof HL): Similar Triangles
If two triangles are equiangular, then the ratio of the lengths of corresponding sides is the same.
Proof:
1. Given: ΔABC and ΔA'B'C' are equiangular (i.e., ∠A = ∠A', ∠B = ∠B', ∠C = ∠C').
2. To Prove: |AB|/|A'B'| = |BC|/|B'C'| = |CA|/|C'A'|.
3. Construction: On [AB], mark a point P such that |AP| = |A'B'|. On [AC], mark a point Q such that |AQ| = |A'C'|. Join P to Q.
4. Proof:
* Consider ΔAPQ and ΔA'B'C'.
* |AP| = |A'B'| (by construction).
* |AQ| = |A'C'| (by construction).
* ∠PAQ = ∠B'A'C' (given ∠A = ∠A').
* Therefore, ΔAPQ is congruent to ΔA'B'C' (SAS).
* From congruence, ∠APQ = ∠A'B'C' and ∠AQP = ∠A'C'B'.
* Given ∠A'B'C' = ∠ABC, it follows that ∠APQ = ∠ABC.
* Since ∠APQ and ∠ABC are corresponding angles and are equal, PQ || BC.
* Now, by Theorem 12 (the ratio theorem for parallel lines in a triangle), we have:
* |AB|/|AP| = |AC|/|AQ|.
* Substituting |AP| = |A'B'| and |AQ| = |A'C'| (from construction):
* |AB|/|A'B'| = |AC|/|A'C'|.
* By repeating the process, placing ΔA'B'C' on ΔABC such that B' coincides with B, we can similarly show:
* |AB|/|A'B'| = |BC|/|B'C'|.
* Combining these results, we conclude: |AB|/|A'B'| = |BC|/|B'C'| = |CA|/|C'A'|.
Theorem 14: Angle at Centre and Circumference
The angle at the centre of a circle standing on a given arc is twice the angle at any point of the circumference standing on the same arc.
Theorem 19: Intersecting Chords Theorem
If two chords AB and CD of a circle intersect at a point P, inside the circle, then the product of the segments of one chord equals the product of the segments of the other chord.
|AP| . |PB| = |CP| . |PD|
Construction 16: Bisector of an Angle
To divide a given angle into two equal angles.
Steps:
1. Place the compass point at the vertex of the angle. Draw an arc that intersects both arms of the angle.
2. From each of these intersection points, draw another arc such that the two new arcs intersect inside the angle.
3. Draw a straight line from the vertex to the intersection point of the two new arcs. This line is the angle bisector.
Construction 17: Perpendicular from a Point to a Line
To construct a line perpendicular to a given line that passes through a given point (either on or off the line).
Steps (Point not on line):
1. Place the compass point at the given point. Draw an arc that intersects the line at two points.
2. From each of these intersection points, draw an arc on the opposite side of the line (or same side, as long as they intersect).
3. Draw a straight line from the given point to the intersection point of the two new arcs. This line is perpendicular to the given line.
Steps (Point on line):
1. Place the compass point at the given point on the line. Draw arcs of equal radius on both sides of the point, intersecting the line.
2. From each of these two intersection points, draw arcs of a larger radius, intersecting above (or below) the line.
3. Draw a straight line from the given point through the intersection of the two arcs. This line is perpendicular to the given line.
Construction 18: Perpendicular Bisector of a Segment
To construct a line that is perpendicular to a given line segment and passes through its midpoint.
Steps:
1. Place the compass point at one endpoint of the segment. Open the compass to more than half the length of the segment. Draw an arc above and below the segment.
2. Without changing the compass width, place the compass point at the other endpoint and draw arcs that intersect the first two arcs.
3. Draw a straight line connecting the two intersection points of the arcs. This line is the perpendicular bisector.
Construction 19: Line Parallel to a Given Line, Through a Given Point
To construct a line that is parallel to a given line and passes through a specific point not on the line.
Steps:
1. Draw a transversal line through the given point and intersecting the given line.
2. At the given point, construct an angle equal to the alternate interior angle (or corresponding angle) formed by the transversal and the given line (using Construction 16 principles).
3. The new line forming this angle will be parallel to the given line.
Construction 20: Division of a Line Segment into n Equal Segments
To divide a line segment into any number of equal parts without measuring it.
Steps:
1. Draw the given line segment AB.
2. From point A, draw a ray AX at any acute angle to AB.
3. Using a compass, mark off n equal segments of any convenient length along AX (e.g., A1, A2, ..., An).
4. Join the last point An to B.
5. Through each of the points A1, A2, ..., An-1, draw lines parallel to AnB. These parallel lines will divide AB into n equal segments (by Theorem 11).
Construction 21: Tangent to a Circle at a Given Point on the Circle
To construct a line that touches the circle at exactly one point (the given point).
Steps:
1. Draw a radius from the centre of the circle to the given point on the circumference.
2. At the given point, construct a line perpendicular to this radius (using Construction 17). This perpendicular line is the tangent.
Construction 22: Tangents to a Circle from an External Point
To construct two lines from a point outside the circle that are tangent to the circle.
Steps:
1. Let O be the centre of the circle and P be the external point.
2. Draw the line segment OP.
3. Construct the perpendicular bisector of OP. Let M be the midpoint of OP (using Construction 18).
4. With M as centre and radius MO (or MP), draw a circle. This circle will intersect the original circle at two points, say T1 and T2.
5. Draw lines PT1 and PT2. These are the two tangents from P to the circle.
Key facts to remember
- 1The proofs for Theorems 11, 12, and 13 are examinable at Leaving Certificate Higher Level.
- 2Theorem 12 is fundamental for understanding similar triangles and proportional division of sides.
- 3Theorem 13 is the definition of similar triangles: equiangular implies proportional sides.
- 4Theorem 19 (Intersecting Chords) is a key result for problems involving chords within a circle.
- 5All constructions must be performed using only a compass and a straightedge.
- 6Many Junior Cycle theorems (e.g., Theorems 4, 6, 9) are foundational and frequently used in Leaving Cert problems.
Worked examples
Example 1
In triangle ABC, a line segment DE is parallel to BC, with D on AB and E on AC. If |AD| = 4 cm, |DB| = 2 cm, and |AE| = 6 cm, find the length of |EC|.
I1. Identify the relevant theorem: Since DE || BC, by Theorem 12 (Proportional Division of Sides in a Triangle), the line DE divides the sides AB and AC proportionally. A common corollary states |AD|/|DB| = |AE|/|EC|.
II2. Substitute the given values into the proportion: 4/2 = 6/|EC|.
III3. Simplify the ratio: 2 = 6/|EC|.
IV4. Solve for |EC|: 2 * |EC| = 6 => |EC| = 6/2 => |EC| = 3 cm.
Answer
|EC| = 3 cm.
This problem demonstrates a direct application of Theorem 12 or its corollary.
Example 2
Chords AB and CD of a circle intersect at point P inside the circle. If |AP| = 6 cm, |PB| = 4 cm, and |CP| = 3 cm, find the length of |PD|.
I1. Identify the relevant theorem: The problem involves two chords intersecting inside a circle, so Theorem 19 (Intersecting Chords Theorem) applies.
II2. State Theorem 19: |AP| . |PB| = |CP| . |PD|.
III3. Substitute the given values into the formula: 6 * 4 = 3 * |PD|.
IV4. Calculate the product: 24 = 3 * |PD|.
V5. Solve for |PD|: |PD| = 24 / 3.
VI6. Result: |PD| = 8 cm.
Answer
|PD| = 8 cm.
Ensure you multiply the segments of each chord together, not add them.
Example 3
Construct a line segment of length 8 cm and divide it into 3 equal parts without measuring.
I1. Draw a line segment AB of length 8 cm.
II2. From point A, draw a ray AX at an acute angle to AB.
III3. Using a compass, mark off 3 equal segments of any convenient length along AX. Label these points A1, A2, A3, starting from A (so AA1 = A1A2 = A2A3).
IV4. Join the last point A3 to point B.
V5. Through each of the points A1 and A2, draw lines parallel to A3B. These parallel lines will intersect AB at points P1 and P2.
VI6. The segment AB is now divided into 3 equal parts: AP1, P1P2, P2B.
Answer
The construction steps describe the division of the 8 cm line segment into 3 equal parts.
This construction relies on Theorem 11, which states that parallel lines cutting equal segments on one transversal will cut equal segments on any other transversal.
Common mistakes
- ✗Confusing corresponding sides when setting up ratios for similar triangles (Theorem 13).
- ✗Incorrectly applying Theorem 12, for example, assuming segments are equal rather than proportional.
- ✗Forgetting the precise sequence of steps for geometric constructions, leading to inaccurate drawings.
- ✗Failing to justify geometric statements with the correct theorem or axiom in proofs and problem-solving.
- ✗Assuming properties (e.g., lines are parallel) without sufficient evidence or explicit statement in the problem.
Exam tips
- ★**Master the Proofs**: Practice writing out the proofs for Theorems 11, 12, and 13 repeatedly until you can reproduce them accurately and efficiently, including 'Given', 'To Prove', 'Construction', and 'Proof' sections.
- ★**Practice Constructions**: Regularly perform all constructions (16-22) using a compass and straightedge. Focus on precision and neatness, as marks are often awarded for clear construction lines.
- ★**Draw Clear Diagrams**: For all geometry problems, draw a large, clear, and well-labelled diagram. This helps visualise the problem and identify which theorems are applicable.
- ★**Justify Every Step**: In solutions and proofs, explicitly state the theorem or axiom that supports each geometric statement you make. For example, 'By Theorem 14...' or 'Vertically opposite angles are equal...'. This is crucial for earning full marks.
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