Mathematics • Year 10 • Unit 3 • Lesson 14
Mixed Geometry — Mixed Challenge
Pull every Lesson 14 idea together: angle-chase with parallel lines, isosceles and parallelogram properties, congruence vs similarity, the midpoint-theorem result, and short formal proofs with statement-plus-reason structure. Spot the classic "I used the wrong tool" mistake, then design your own multi-step angle-chase problem.
1. Mixed problems
Each problem combines at least two of the Lesson 14 ideas. Mark the diagram with what you know before you start. 2-3 marks each
1.1 △ABC is isosceles with AB = AC and ∠BAC = 40°. D is on BC so that AD ⊥ BC. Find ∠BAD and prove △ABD ≡ △ACD. 3 marks
1.2 In parallelogram ABCD, AB ∥ CD and AD ∥ BC. Prove that △ABD ≡ △CDB. (Hint: BD is common.) 3 marks
1.3 Two parallel lines are cut by a transversal so that one of the co-interior angles is x° and the other is (3x − 20)°. Find x. 2 marks
1.4 In △ABC, D is the midpoint of AB and E is the midpoint of AC. Use a similarity argument to prove DE = (1/2) × BC. 3 marks
1.5 A quadrilateral ABCD has all four sides equal (AB = BC = CD = DA) and one angle of 60°. Find all four angles, and name the shape. 3 marks
1.6 △ABC has ∠A = 40° and ∠B = 70°. Point D is on AB and point E is on AC such that DE ∥ BC. Find all three angles of △ADE. 2 marks
2. Find the mistake
Another Year 10 student has tried to prove two triangles are congruent. Their working is shown. Exactly one line contains a mistake. Spot it, explain why it is wrong, and re-do the working correctly. 3 marks
Setup: △ABC and △DEF have AB = DE = 6 cm, AC = DF = 8 cm, and ∠B = ∠E = 40°. Student's working:
Line 1: AB = DE = 6 (given)
Line 2: AC = DF = 8 (given)
Line 3: ∠B = ∠E = 40° (given) — this is the included angle between the two equal sides.
Line 4: Therefore △ABC ≡ △DEF (SAS).
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) What extra information would actually be needed to apply SAS legitimately? (Or, state which test would actually be valid here.)
Stuck? In △ABC, the side AC is opposite ∠B — not adjacent to it. The included angle between AB and AC is ∠A, not ∠B.3. Open-ended challenge — design your own angle chase
This question has many valid answers. Be creative but show every reason. 4 marks
3.1 Design a figure that combines at least three of the following geometric facts:
- parallel lines and a transversal,
- an isosceles triangle,
- a parallelogram or rhombus,
- two triangles that are either congruent or similar,
- the 180° angle sum in a triangle.
In your answer:
(i) Sketch the figure and mark all the given angles or equal sides.
(ii) State one unknown angle in the figure.
(iii) Find the unknown angle, writing every step as statement + reason.
(iv) State which facts from Lesson 14 you used.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Isosceles with perpendicular from apex
Base angles ∠ABC = ∠ACB = (180° − 40°) / 2 = 70°.
In △ABD: ∠ADB = 90°, ∠ABD = 70°, so ∠BAD = 180° − 90° − 70° = 20°.
Proof:
AB = AC (given).
∠ABD = ∠ACD = 70° (base angles equal).
∠ADB = ∠ADC = 90° (AD ⊥ BC).
Therefore △ABD ≡ △ACD (AAS). (Equivalently RHS, since AD is common and AB = AC are the hypotenuses.)
1.2 — Parallelogram △ABD ≡ △CDB
AB = CD (opposite sides of parallelogram are equal).
AD = CB (opposite sides of parallelogram are equal).
BD = BD (common side).
Therefore △ABD ≡ △CDB (SSS).
1.3 — Co-interior angles in terms of x
Co-interior angles between parallel lines sum to 180°:
x + (3x − 20) = 180.
4x = 200, so x = 50°.
1.4 — Midpoint theorem via similarity
∠A is common to △ADE and △ABC.
AD/AB = 1/2 (D is midpoint of AB). AE/AC = 1/2 (E is midpoint of AC).
Therefore △ADE ~ △ABC (SAS similarity) with scale factor 1/2.
Hence DE = (1/2) × BC. (And ∠ADE = ∠ABC, so DE ∥ BC.)
1.5 — Rhombus with a 60° angle
All sides equal → rhombus. Adjacent angles are co-interior, so they sum to 180°.
If one angle is 60°, the adjacent angle is 120°. Opposite angles are equal.
Angles: 60°, 120°, 60°, 120°.
1.6 — △ADE with DE ∥ BC
△ADE ~ △ABC by AAA, so the angles match.
∠A = 40° (common). ∠ADE = ∠ABC = 70° (corresponding angles, DE ∥ BC).
∠AED = 180° − 40° − 70° = 70°.
So △ADE has angles 40°, 70°, 70°.
2 — Find the mistake
(a) The mistake is on Line 3 (which then makes Line 4 invalid).
(b) ∠B is the angle at vertex B, which lies between sides AB and BC — not between AB and AC. The included angle between AB and AC is ∠A. So the equal angles ∠B = ∠E are not the included angles between the two given pairs of equal sides, and SAS does not apply.
(c) To apply SAS we would need ∠A = ∠D (the angle included between AB and AC). Alternatively, since AB = DE = 6, ∠B = ∠E and AC = DF = 8 (a non-included angle), this is the ambiguous SSA case for congruence and the proof shown is invalid as stated.
3 — Open-ended (sample solution)
Figure: A parallelogram ABCD with AB ∥ CD. Inside, a diagonal AC is drawn, and along that diagonal the triangle △ACD is identified. Given: ∠DAB = 110°, AD = DC (so △ACD is isosceles).
Unknown: ∠DCA.
Working:
∠ADC = 180° − 110° = 70° (co-interior angles between parallel lines AB ∥ CD).
∠DAC = ∠DCA (base angles of isosceles △ACD, since AD = DC).
In △ACD: ∠DAC + ∠DCA + ∠ADC = 180° → 2 × ∠DCA + 70° = 180° → ∠DCA = 55°.
Facts used: co-interior angles between parallel sides of a parallelogram, isosceles triangle base-angle property, 180° triangle angle sum.
Marking: 1 for a sensible figure combining at least three facts, 1 for choosing and clearly stating an unknown, 1 for correct statement-plus-reason working, 1 for correctly naming the Lesson 14 facts used. Any valid figure earning the unknown by valid reasoning is acceptable.