Mathematics • Year 10 • Unit 3 • Lesson 14
Mixed Geometry Problems — Angle-Chase Skill Drill
Build fluency with multi-step angle chases and short congruence / similarity proofs. Each problem combines parallel-line angles, isosceles-triangle properties, the 180° triangle sum, and the right tool (congruence or similarity) for the job — exactly the toolkit Lesson 14 mixes.
1. I do — fully worked angle chase
Read each step. Every line is a statement + reason. That is exactly how a Year 10 geometry proof reads.
Problem. In △ABC, AB = AC and ∠BAC = 40°. Find ∠ABC.
Step 1 — Identify the triangle's properties.
△ABC is isosceles with AB = AC.
Reason: given.
Step 2 — Use the equal-sides → equal-angles property.
∠ABC = ∠ACB
Reason: base angles of an isosceles triangle are equal.
Step 3 — Apply the 180° angle-sum rule.
∠ABC + ∠ACB + ∠BAC = 180°
2 × ∠ABC + 40° = 180°
Reason: angles in a triangle sum to 180°; the two base angles are equal.
Step 4 — Solve for ∠ABC.
∠ABC = (180° − 40°) / 2 = 70°
Answer: ∠ABC = 70°.
2. We do — fill in the missing steps
Fill in each blank. 5 marks
Problem. Two parallel lines are cut by a transversal. One of the angles formed is 65°. Find the alternate angle, the corresponding angle, and the co-interior angle on the other side.
Step 1 — Alternate angle:
Alternate angle = ______° (reason: __________________________)
Step 2 — Corresponding angle:
Corresponding angle = ______° (reason: __________________________)
Step 3 — Co-interior angle (same side of the transversal):
Co-interior angle = 180° − ______ = ______° (reason: __________________________)
Step 4 — Sanity-check: two adjacent angles on a straight line sum to ______°.
3. You do — independent practice
Show your working. Foundation (3.1-3.4), standard (3.5-3.6), extension (3.7-3.8).
Foundation — single-step angle chase
3.1 In an isosceles triangle, the apex angle is 50°. Find each base angle. 1 mark
3.2 Two angles of a triangle are 35° and 95°. Find the third. 1 mark
3.3 In a parallelogram, one angle is 110°. Find the other three. 1 mark
3.4 A quadrilateral has diagonals that bisect each other. What type of quadrilateral must it be? Give the reason. 1 mark
Standard — two-step angle chase
3.5 △ABC is isosceles with AB = AC and ∠BAC = 80°. D is the midpoint of BC. Find ∠BAD and ∠ADB. 2 marks
3.6 △ABC and △DEF have ∠A = ∠D = 50°, ∠B = ∠E = 60°. AB = 4 cm, DE = 8 cm. Find the scale factor and the length of EF if BC = 5 cm. 2 marks
Extension — short proof
3.7 In △ABC, AB = AC. M is the midpoint of BC, and AM is drawn. Prove that △ABM ≡ △ACM. State the test used. 3 marks
3.8 In △ABC, D is the midpoint of AB and E is the midpoint of AC. Prove that △ADE ~ △ABC, and state the scale factor. (Hint: use SAS similarity with ∠A common.) 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (parallel lines, 65°)
Step 1: Alternate angle = 65° (alternate angles between parallel lines are equal).
Step 2: Corresponding angle = 65° (corresponding angles between parallel lines are equal).
Step 3: Co-interior angle = 180° − 65° = 115° (co-interior angles between parallel lines are supplementary).
Step 4: 180° (angles on a straight line).
3.1 — Apex 50°
Each base angle = (180° − 50°) / 2 = 65°.
3.2 — Two angles 35° and 95°
Third angle = 180° − 35° − 95° = 50°.
3.3 — Parallelogram angle 110°
Opposite angle = 110°. The other two angles = 180° − 110° = 70° each (co-interior angles between the parallel sides).
3.4 — Diagonals bisect each other
It must be a parallelogram (this is the defining converse property: a quadrilateral whose diagonals bisect each other is a parallelogram).
3.5 — Isosceles, apex 80°, D midpoint
By symmetry, AD bisects the apex angle, so ∠BAD = 80° / 2 = 40°. AD also bisects BC at right angles in an isosceles triangle, so ∠ADB = 90°. Check: in △ABD, 40° + 90° + ∠ABD = 180° → ∠ABD = 50° (= (180−80)/2 ✓).
3.6 — Similar triangles, scale factor + EF
k = DE / AB = 8 / 4 = 2. EF = BC × k = 5 × 2 = 10 cm.
3.7 — △ABM ≡ △ACM (proof)
AB = AC (given).
BM = CM (M is the midpoint of BC).
AM = AM (common side).
Therefore △ABM ≡ △ACM (SSS).
3.8 — △ADE ~ △ABC (proof)
∠DAE = ∠BAC (common angle, ∠A).
AD / AB = 1/2 (D is the midpoint of AB).
AE / AC = 1/2 (E is the midpoint of AC).
So two pairs of sides are in ratio 1:2 and the included angle is equal.
Therefore △ADE ~ △ABC (SAS similarity), with scale factor 1/2.