The Lumenary · Grades 9–12

Euclidean geometry and proof

Name: ______________________

Date: ______________________

You will learn to

Use the congruence criteria SAS, ASA and SSS to prove properties of triangles and quadrilaterals, and explain why SSA is not a criterion
Prove, using the parallel postulate, that the angles of a triangle sum to $180^\circ$
Prove the inscribed angle theorem and its corollaries, including Thales' theorem and cyclic quadrilaterals
Prove that the perpendicular bisectors of the sides of a triangle meet at its circumcentre
Explain the role of the parallel postulate and how its independence led to non-Euclidean geometries

1.
Two angles of a triangle measure $47^\circ$ and $68^\circ$ . How many degrees is the third angle?
2.
In triangle $ABC$ , $\angle A = 50^\circ$ and $\angle B = 75^\circ$ . The side $BC$ is extended beyond $C$ to a point $D$ . How many degrees is the exterior angle $\angle ACD$ ?
3.
Which of these always guarantee that two triangles are congruent? (Choose all that apply.)
- A.Two sides and the included angle equal (SAS)
- B.All three sides equal (SSS)
- C.Two angles and the side between them equal (ASA)
- D.Two sides and a non-included angle equal (SSA)
- E.All three angles equal (AAA)
4.
In the proof that the angles of a triangle sum to $180^\circ$ , we draw the line through $C$ parallel to $AB$ and say that the angle it makes with $CA$ equals $\angle A$ . Which assumption is essential for that step?
- A.Playfair's axiom (the parallel postulate)
- B.The SAS axiom
- C.Euclid I.27: equal alternate angles imply parallel lines
- D.Thales' theorem
5.
Points $A$ , $B$ , $C$ lie on a circle with centre $O$ , and $B$ is on the same side of $AC$ as $O$ . The central angle $\angle AOC$ is $110^\circ$ . How many degrees is the inscribed angle $\angle ABC$ ?
6.
The vertices of quadrilateral $PQRS$ lie on a circle, in this order, and $\angle PQR = 73^\circ$ . How many degrees is $\angle RSP$ ?
7.
How many degrees is each interior angle of a regular polygon with $12$ sides?
8.
The fallacious proof that every triangle is isosceles lets the bisector of the angle at $A$ meet the perpendicular bisector of $BC$ at a point $P$ , and draws $P$ inside the triangle. Where does $P$ really lie when $AB \ne AC$ ?
- A.On the circumcircle, at the midpoint of the arc $BC$ not containing $A$
- B.At the incentre, inside the triangle
- C.At the circumcentre
- D.Nowhere: the two lines are parallel
9.
Which Lean statement correctly says that SSA is not a congruence criterion? (Here ≅ means the triangles are congruent.)
- A.¬ ∀ A B C A' B' C', ∠ B A C = ∠ B' A' C' → dist A B = dist A' B' → dist B C = dist B' C' → ![A, B, C] ≅ ![A', B', C']
- B.∀ A B C A' B' C', ∠ B A C = ∠ B' A' C' → dist A B = dist A' B' → dist B C = dist B' C' → ¬ (![A, B, C] ≅ ![A', B', C'])
- C.∃ A B C A' B' C', ∠ B A C = ∠ B' A' C' ∧ ![A, B, C] ≅ ![A', B', C']
- D.∀ A B C C', dist A C ≠ dist A C'
10.
Write circumcentre(a, b, c) for points given as pairs of integers. It should return the circumcentre as a pair of Fractions, or None if the points are collinear. Use the lemma: $O$ is equidistant from $A$ and $B$ exactly when $2(B - A) \cdot O = |B|^2 - |A|^2$ . Two such linear equations (for $AB$ and $AC$ ) determine $O$ .
Computational — Practice

Answer key — Euclidean geometry and proof

1.
By the angle sum theorem, the third angle is $180^\circ - 47^\circ - 68^\circ = 65^\circ$ .
2.
By the exterior angle theorem, $\angle ACD = \angle A + \angle B = 50^\circ + 75^\circ = 125^\circ$ . Check: $\angle ACB = 180^\circ - 125^\circ = 55^\circ$ , and $55^\circ + 125^\circ = 180^\circ$ on the straight line $BD$ .
3.
SAS, SSS and ASA are congruence criteria. SSA is not (the circle around $B$ can meet the ray from $A$ twice), and AAA only gives similarity.
4.
The step uses the alternate angles theorem, whose proof needs Playfair's axiom: the line through $C$ making equal alternate angles is parallel to $AB$ (by I.27), and since there is only one parallel through $C$ , it is our line. Without the parallel postulate the angle sum can differ from $180^\circ$ .
5.
By the inscribed angle theorem, $\angle ABC = \tfrac12 \angle AOC = 55^\circ$ .
6.
Opposite angles of a cyclic quadrilateral are supplementary, so $\angle RSP = 180^\circ - 73^\circ = 107^\circ$ .
7.
The angle sum is $(12 - 2) \cdot 180^\circ = 1800^\circ$ , so each angle is $1800^\circ / 12 = 150^\circ$ . (Alternatively: each exterior angle is $360^\circ / 12 = 30^\circ$ , and $180^\circ - 30^\circ = 150^\circ$ .)
8.
Let the bisector of $\angle A$ meet the circumcircle again at $M$ . The inscribed angles $\angle BAM$ and $\angle MAC$ are equal, so they stand on equal arcs, hence equal chords: $MB = MC$ . So $M$ is also on the perpendicular bisector of $BC$ , and $P = M$ lies on the circumcircle, outside the triangle. The diagram in the 'proof' was simply wrong.
9.
'SSA is not a criterion' is the negation of 'for all triangles, SSA data imply congruence', so the ¬ goes in front of the ∀. It is proved by one counterexample. Putting ¬ in front of the conclusion would claim that SSA data never give congruent triangles, which is false.
10.
```
from fractions import Fraction

def circumcentre(a, b, c):
    (ax, ay), (bx, by), (cx, cy) = a, b, c
    a1, b1 = 2 * (bx - ax), 2 * (by - ay)
    c1 = bx ** 2 + by ** 2 - ax ** 2 - ay ** 2
    a2, b2 = 2 * (cx - ax), 2 * (cy - ay)
    c2 = cx ** 2 + cy ** 2 - ax ** 2 - ay ** 2
    det = a1 * b2 - a2 * b1
    if det == 0:
        return None
    return (Fraction(c1 * b2 - c2 * b1, det), Fraction(a1 * c2 - a2 * c1, det))
```
Expanding $|O - A|^2 = |O - B|^2$ gives $2(B - A)\cdot O = |B|^2 - |A|^2$ : the perpendicular bisector of $AB$ as a linear equation. The determinant $D$ vanishes exactly when $B - A$ and $C - A$ are parallel, that is, when the points are collinear — matching the proof that two perpendicular bisectors of a genuine triangle must meet. For the right triangle the answer is the midpoint of the hypotenuse, as Thales' theorem predicts.