L26 Solutions — Geometric Proofs

1. Angle facts. Fluency

   • (a) (3x+10)+(2x−5)=180 ⇒ 5x+5=180 ⇒ x = 35.
   • (b) y = 360−120−95 = 145°.
   • (c) ∠BOC = 180−65 = 115° (angles on a straight line).
   • (d) Other angle = 118−54 = 64° (exterior angle = sum of two non-adjacent interior angles).

2. Parallel lines. Fluency

   • (a) 47° (alternate angles are equal).
   • (b) (4x+15)+(3x+20)=180 ⇒ 7x+35=180 ⇒ x = 20.71… ≈ 20.7. (Check: 7x=145, x=145/7.)
   • (c) 5y−10 = 3y+14 ⇒ 2y=24 ⇒ y = 12.
   • (d) No. 73° ≠ 74°; alternate angles must be equal for lines to be parallel.

3. Triangle proofs. Fluency

   • (a) ∠C = 180−55−70 = 55° (angle sum of triangle = 180°).
   • (b) ∠R = ∠Q = 38° (base angles of isosceles). ∠P = 180−38−38 = 104°.
   • (c) Each interior angle = 60°. Exterior angle = 180−60 = 120° (angles on a straight line).
   • (d) If ∠A = ∠B, the sides opposite these angles are equal: BC = AC (converse of isosceles triangle theorem).

4. Proof of congruence. Fluency

   • (a) AB=DE, BC=EF, ∠ABC=∠DEF (included). SAS: ▵ABC ≅ ▵DEF.
   • (b) ∠P=∠X, ∠Q=∠Y, PQ=XY (corresponding side). AAS: ▵PQR ≅ ▵XYZ.
   • (c) Right angle + hypotenuse 13 + side 5: RHS: congruent.
   • (d) All three sides equal: SSS: ▵ABC ≅ ▵DEF.

5. Parallel lines proof (diagram). Understanding

   • (a) a = 65°. Alternate angles are equal (PQ ∥ RS).
   • (b) b = 180−65 = 115°. Co-interior angles sum to 180° (PQ ∥ RS).
   • (c) c = b = 115°. Vertically opposite angles are equal.
   • (d) d = 50°. Corresponding angles are equal (PQ ∥ RS).

6. Rhombus diagonals bisect at right angles. Understanding

   • (a) In ▵AOB and ▵COB: AB = CB (rhombus, all sides equal); OB = OB (common); OA = OC (diagonals bisect each other in a parallelogram). ∴ ▵AOB ≅ ▵COB (SSS).
   • (b) ∠AOB = ∠COB (corresponding angles, ▵AOB ≅ ▵COB). ∠AOB + ∠COB = 180° (angles on straight line AC). ∴ 2∠AOB = 180° ⇒ ∠AOB = 90°.
   • (c) In a parallelogram (all rhombuses are parallelograms), diagonals bisect each other: AO = CO and BO = DO (property of parallelogram, or prove by congruent triangles ▵AOB ≅ ▵COD by ASA).
   • (d) In a rhombus, diagonals are perpendicular bisectors of each other. In a rectangle, diagonals are equal in length but not necessarily perpendicular.

7. Two-column proofs. Understanding

   • (a) ∠AOC + ∠BOC = 180° (straight line AB). ∠BOD + ∠BOC = 180° (straight line CD). ∴ ∠AOC = 180°−∠BOC = ∠BOD. Vertically opposite angles are equal. □
   • (b) ∠A + ∠B + ∠ACB = 180° (triangle angle sum). ∠ACB + ∠ACD = 180° (straight line). ∴ ∠ACD = ∠A + ∠B. □
   • (c) ▵ABD ≅ ▵ACD (SSS: AB=AC given, BD=DC since D midpoint, AD common). ∴ ∠ABD = ∠ACD (corresponding angles). □
   • (d) Alternate angles = 52° = 52° (equal). By the converse of the alternate angles theorem, l₁ ∥ l₂. □

8. Mid-point theorem. Understanding

   • (a) ∠AMN = ∠ABC (same angle at A). AM/AB = AN/AC = 1/2 (M, N midpoints). ∴ ▵AMN ~ ▵ABC (SAS similarity).
   • (b) ∠AMN = ∠ABC (from similarity). These are corresponding angles with transversal AB. ∴ MN ∥ BC. □
   • (c) Scale factor = AM/AB = 1/2. MN = (1/2)BC. □
   • (d) MN = 14/2 = 7 cm. AB = 2×AM = 2×5 = 10 cm.

9. Angle chase (AB ∥ CD). Problem Solving

   • (a) ∠EPQ = 180−35 = 145° (angles on straight line AB at P). Or: ∠EPQ below AB = 35° if E is above; depends on configuration. Treating ∠APE = 35° as above AB: angle on the other side = 180−35 = 145°. The interior angle at P in ▵PQR = 180−35 = 145° or 35° depending on orientation. Taking ∠QPR (interior) = 35° (alternate to ∠APE if PQ is the transversal) — see part (d).
   • (b) ∠FQP (interior angle at Q in ▵PQR) = 48° (alternate to ∠DQF, since AB ∥ CD).
   • (c) ∠PRQ = 180−35−48 = 97° (angle sum of ▵PQR).
   • (d) ∠PRQ = ∠APE + ∠DQF = 35 + 48 = 83°... wait: by the exterior angle theorem, ∠PRQ = ∠QPR + ∠PQR = 35 + 48 = 83°. But we computed 97° above; this means ∠PRQ itself is the angle at R, which is 180−35−48 = 97°. The relationship ∠PRQ = ∠APE + ∠DQF = 83° refers to the alternate interpretation. With AB ∥ CD and R between the lines: ∠PRQ = |∠APE + ∠DQF| = 83° or 97° depending on which side R falls.

10. Pythagoras’ theorem (geometric proof). Problem Solving

    • (a) Large square area = (a+b)². Also = 4 triangles + central square = 4×(½ab) + c² = 2ab + c².
    • (b) (a+b)² = 2ab + c² ⇒ a² + 2ab + b² = 2ab + c². This is what we need to show.
    • (c) Cancelling 2ab from both sides: a² + b² = c². □
    • (d) Yes, valid. Assumptions: the central quadrilateral is indeed a square (requires showing all angles are 90°, which follows since the four triangles are right-angled and congruent), and the large square has equal sides (a+b).